Appendix: Proof of Theorem 1
Since μ(I
00, …, I
pq
) is a combination of blur invariants I
ij
only, it is also invariant to blur distortion. To validate its invariance to affine transformation, we need to prove that
$$ \frac{{\partial \mu (I_{00} , \ldots ,I_{pq} )}}{\partial a} \equiv \sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {\frac{{\partial \mu (I_{00} , \ldots ,I_{pq} )}}{{\partial I_{ij} }}} } \frac{{\partial I_{ij} }}{\partial a} = 0 $$
(25)
for each parameter a
\( \in \){a
11, a
12, a
21, a
22, b
11, b
12} of the affine transform.
As μ(L
00, …, L
pq
) is invariant to affine transform, we have
$$ \frac{{\partial \mu (L_{00} , \ldots ,L_{pq} )}}{\partial a} \equiv \sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {\frac{{\partial \mu (L_{00} , \ldots ,L_{pq} )}}{{\partial L_{ij} }}} } \frac{{\partial L_{ij} }}{\partial a} = 0 $$
(26)
Thus, to verify the affine invariance of μ(I
00,…, I
pq
), it is sufficient to prove
$$ \frac{{\partial I_{ij} }}{\partial a} = \frac{{\partial L_{ij} }}{\partial a}\left| {_{I = L} } \right. $$
(27)
As the affine transform can be decomposed into four transformations which are described in Sect. 3.1, we need to prove that (27) holds for each of these transformations. Obviously, (27) holds for translation in (12). Therefore, we only need to validate the other three transformations.
1.1 A. X-shearing
For X-shearing transformation, we have
$$ \frac{{\partial L_{pq} }}{\partial \beta } = \frac{\partial }{\partial \beta }\int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {P_{p} (u)P_{q} (v)\ddot{f}(u,v)dudv} } = \frac{\partial }{\partial \beta }\int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {P_{p} (x + \beta y)P_{q} (y)f(x,y)dxdy} } $$
(28)
Substitution of (2) into (28) leads to
$$ \begin{aligned} \frac{{\partial L_{pq} }}{\partial \beta } & = \frac{\partial }{\partial \beta }\int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {\sum\limits_{m = 0}^{p} {c_{p,m} (x + \beta y)^{m} \sum\limits_{n = 0}^{q} {c_{q,n} y^{n} f(x,y){\text{d}}x{\text{d}}y} } } } \\ & = \int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {\sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {c_{p,m} c_{q,n} m(x + Ty)^{m - 1} y^{n + 1} f(x,y){\text{d}}x{\text{d}}y} } } } \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {c_{p,m} c_{q,n} m\int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {u^{m - 1} v^{n + 1} \ddot{f}(u,v){\text{d}}u{\text{d}}v} } } } \\ \end{aligned} $$
(29)
Using (5), (29) can be rewritten as
$$ \begin{aligned} \frac{{\partial L_{pq} }}{\partial \beta } & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {c_{p,m} c_{q,n} m\int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {\sum\limits_{i = 0}^{m - 1} {d_{m - 1,i} P_{i} (u)\sum\limits_{j = 0}^{n + 1} {d_{n + 1,j} P_{j} (v)} } \ddot{f}(u,v){\text{d}}u{\text{d}}v} } } } \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {mc_{p,m} c_{q,n} \sum\limits_{i = 0}^{m - 1} {\sum\limits_{j = 0}^{n + 1} {d_{m - 1,i} d_{n + 1,j} } } \int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {P_{i} (u)P_{j} (v)\ddot{f}(u,v){\text{d}}u{\text{d}}v} } } } \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{i = 0}^{m - 1} {\sum\limits_{j = 0}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,i} d_{n + 1,j} L_{ij} } } } } \\ \end{aligned} $$
(30)
In order to validate (27) for X-shearing transformation, we need to prove
$$ \frac{{\partial I_{pq} }}{\partial \beta } = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{i = 0}^{m - 1} {\sum\limits_{j = 0}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,i} d_{n + 1,j} I_{ij} } } } } $$
(31)
In the next procedures, we will use induction principle to prove (31). It can be easily verified that (31) is true for p + q = 1, because in that case I
pq
is equal to L
pq
. For p + q = 3, we have four cases to be considered: p = 3, q = 0; p = 2, q = 1; p = 1, q = 2; p = 0, q = 3. Now the proof for p = 2, q = 1 is given as follows, while other cases can be validated in a similar way.
From the definition of blur invariants given in (10) and (30), we have
$$ \begin{aligned} \frac{{\partial I_{21} }}{\partial \beta } = & \sum\limits_{m = 0}^{2} {\sum\limits_{n = 0}^{1} {\sum\limits_{i = 0}^{m - 1} {\sum\limits_{j = 0}^{n + 1} {mc_{2,m} c_{1,n} d_{m - 1,i} d_{n + 1,j} L_{ij} } } } } \\ & - \frac{1}{{2L_{00} }}\mathop{\sum\limits_{i = 0}^{2} {\sum\limits_{j = 0}^{1} {} } }\limits_{0 <i + j< 3} \sum\limits_{s = 0}^{2 - i} \sum\limits_{t = 0}^{1 - j} A(2,1,i,j,s,t)\\&\times\left[ {\frac{\partial I(i,j)}{\partial \beta }L_{st} + I(i,j)\frac{{\partial L_{st} }}{\partial \beta }} \right] \\ \end{aligned} $$
(32)
To prove that (31) holds for p = 2, q = 1, it is sufficient to prove
$$ \frac{{\partial I_{21} }}{\partial \beta } - \sum\limits_{m = 0}^{2} {\sum\limits_{n = 0}^{1} {\sum\limits_{i = 0}^{m - 1} {\sum\limits_{j = 0}^{n + 1} {mc_{2,m} c_{1,n} d_{m - 1,i} d_{n + 1,j} I_{ij} } } } } = 0 $$
(33)
Based on Theorem 1 and (30), we have
$$ \begin{aligned} \frac{{\partial L_{00} }}{\partial \beta } & = 0,\;\frac{{\partial L_{10} }}{\partial \beta } = c_{11} c_{00} d_{00} d_{10} L_{00} + c_{11} c_{00} d_{00} d_{11} L_{01} ,\;\frac{{\partial L_{01} }}{\partial \beta } = 0 \\ I_{12} & = L_{12} - \frac{1}{{2L_{00} }}\{ I_{01} [L_{00} A(1,2,0,1,0,0) + L_{01} A(1,2,0,1,0,1) + L_{10} A(1,2,0,1,1,0) \\ & \quad + L_{11} A(1,2,0,1,1,1)] + I_{10} [L_{00} A(1,2,1,0,0,0) + L_{01} A(1,2,1,0,0,1) + L_{02} A(1,2,1,0,0,2)\} \\ \end{aligned} $$
(34)
$$ L_{00} = I_{00} ,\;L_{01} = I_{01} ,\;L_{10} = I_{10} $$
Substituting (30) and (34) into (33) leads to
$$ \frac{{\partial I_{21} }}{\partial \beta } = \sum\limits_{m = 0}^{2} {\sum\limits_{n = 0}^{1} {\sum\limits_{i = 0}^{m - 1} {\sum\limits_{j = 0}^{n + 1} {mc_{2,m} c_{1,n} d_{m - 1,i} d_{n + 1,j} I_{ij} } } } } $$
(35)
Then, the case of p = 2, q = 1 is proved.
Assume that (31) holds for all invariants of order less than p + q, next we will consider the case of p + q.
Using Theorem 1, we can achieve
$$ \begin{aligned} \frac{{\partial I_{pq} }}{\partial \beta } &= \frac{{\partial L_{p,q} }}{\partial \beta } - \frac{1}{{2L_{00} }}\mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\left[ {\frac{{\partial (L_{st} )}}{\partial \beta }I_{ij} + L_{s,t} \frac{{\partial I_{ij} }}{\partial \beta }} \right] \cdot } A(p,q,i,j,s,t)} \\ &= \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{\omega = 0}^{m - 1} {\sum\limits_{\varepsilon = 0}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,\omega } d_{n + 1,\varepsilon } L_{\omega \varepsilon } } } } }\\&\quad - \frac{1}{{2L_{00} }}\mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q}\\&\quad \times \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\left[ {\sum\limits_{g = 0}^{s} {\sum\limits_{h = 0}^{t} {\sum\limits_{\tau = 0}^{g - 1} {\sum\limits_{\rho = 0}^{h + 1} {gc_{sg} c_{th} d_{g - 1,\tau } d_{h + 1,\rho } L_{\tau \rho } } } } } I_{ij} } \right.} } \\ &\quad \left. { + L_{s,t} \sum\limits_{\lambda = 0}^{i} {\sum\limits_{\sigma = 0}^{j} {\sum\limits_{w = 0}^{\lambda - 1} {\sum\limits_{b = 0}^{\sigma + 1} {\lambda c_{i,\lambda } c_{j,\sigma } d_{\lambda - 1,w} d_{\sigma + 1,b} I_{wb} } } } } } \right] \cdot A(p,q,i,j,s,t) \\ \end{aligned} $$
(36)
Let
$$ \frac{{\partial \hat{I}_{pq} }}{\partial \beta } = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{i = 0}^{m - 1} {\sum\limits_{j = 0}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,i} d_{n + 1,j} I_{ij} } } } } $$
(37)
If we can prove \( \frac{{\partial I_{pq} }}{\partial \beta } - \frac{{\partial \hat{I}_{pq} }}{\partial \beta } = 0 \), then (31) is true for the case of p + q. From Theorem 1, we can deduced that
$$ \begin{aligned} \frac{{\partial \hat{I}_{pq} }}{\partial \beta } & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{e = 0}^{m - 1} {\sum\limits_{r = 0}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,e} d_{n + 1,r} } } } } L_{er} \\ & \;\;\;\; - \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{e = 0}^{m - 1} {\sum\limits_{r = 0}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,e} d_{n + 1,r} } } } } \frac{1}{{2L_{00} }}\mathop {\sum\limits_{i = 0}^{e} {\sum\limits_{j = 0}^{r} {} } }\limits_{0 < i + j < e + r} I_{ij} \sum\limits_{s = 0}^{e - i} {\sum\limits_{t = 0}^{r - j} {L_{s,t} } A(e,r,i,j,s,t)} \\ \end{aligned} $$
(38)
Thus
$$ \begin{aligned} \frac{{\partial \hat{I}_{{pq}} }}{{\partial \beta }} - \frac{{\partial I_{{pq}} }}{{\partial \beta }} & = \frac{1}{{2L_{{00}} }}\mathop {\sum\limits_{{i = 0}}^{p} {\sum\limits_{{j = 0}}^{q} {} } }\limits_{{0 < i + j < p + q}} \sum\limits_{{s = 0}}^{{p - i}} {\sum\limits_{{t = 0}}^{{q - j}} {[\sum\limits_{{g = 0}}^{s} {\sum\limits_{{h = 0}}^{t} {\sum\limits_{{\tau = 0}}^{{g - 1}} {\sum\limits_{{\rho = 0}}^{{h + 1}} {gc_{{s,g}} c_{{t,h}} d_{{g - 1,\tau }} d_{{h + 1,\rho }} L_{{\tau \rho }} } } } } I_{{ij}} } } \\ & + L_{{s,t}} \sum\limits_{{\lambda = 0}}^{i} {\sum\limits_{{\sigma = 0}}^{j} {\sum\limits_{{w = 0}}^{{\lambda - 1}} {\sum\limits_{{b = 0}}^{{\sigma + 1}} {\lambda c_{{i,\lambda }} c_{{j,\sigma }} d_{{\lambda - 1,w}} d_{{\sigma + 1,b}} I_{{wb}} ] \cdot } } } } A(p,q,i,j,s,t) \\ & - \sum\limits_{{m = 0}}^{p} {\sum\limits_{{n = 0}}^{q} {\sum\limits_{{e = 0}}^{{m - 1}} {\sum\limits_{{r = 0}}^{{n + 1}} {mc_{{pm}} c_{{qn}} d_{{m - 1,e}} d_{{n + 1,r}} } } } } \frac{1}{{2L_{{00}} }}\mathop {\sum\limits_{{i = 0}}^{e} {\sum\limits_{{j = 0}}^{r} {} } }\limits_{{0 < i + j < e + r}} I_{{ij}} \sum\limits_{{s = 0}}^{{e - i}} {\sum\limits_{{t = 0}}^{{r - j}} {L_{{s,t}} } A(e,r,i,j,s,t)} \\ \end{aligned} $$
(39)
Define the first item of right side in (39) as follows:
$$ A = \mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\sum\limits_{g = 0}^{s} {\sum\limits_{h = 0}^{t} {\sum\limits_{\tau = 0}^{g - 1} {\sum\limits_{\rho = 0}^{h + 1} {gc_{s,g} c_{t,h} d_{g - 1,\tau } d_{h + 1,\rho } L_{\tau \rho } } } } } I_{ij} A(p,q,i,j,s,t)} } $$
(40)
Let p = p + 1 and q = q− 1, we can rewrite A as:
$$ A = \mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {\left[ {\sum\limits_{g = 0}^{s} {\sum\limits_{h = 0}^{t} {\sum\limits_{\tau = 0}^{g - 1} {\sum\limits_{\rho = 0}^{h + 1} {gc_{s,g} c_{t,h} d_{g - 1,\tau } d_{h + 1,\rho } L_{\tau \rho } A(p - 1,q + 1,i,j,s,t)} } } } } \right.} }, $$
and we also give that
$$ A^{'} = \mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {L_{st} \sum\limits_{\tau = s + 1}^{p - i} {\sum\limits_{\rho = t - 1}^{q - j} {\sum\limits_{g = s + 1}^{\tau } {\sum\limits_{h = t - 1}^{\rho } {gc_{\tau ,g} c_{\rho ,h} d_{g - 1,s} d_{h + 1,t} A(p,q,i,j,\tau ,\rho )} } } } } } $$
(41)
Then
$$ \begin{aligned} A - A^{\prime } & = \mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {\sum\limits_{g = 0}^{s} {\sum\limits_{h = 0}^{t} {\sum\limits_{\tau = 0}^{g - 1} {\sum\limits_{\rho = 0}^{h + 1} {gc_{s,g} c_{t,h} d_{g - 1,\tau } d_{h + 1,\rho } L_{\tau \rho } A(p - 1,q + 1,i,j,s,t)} } } } } } \\ \, & - \mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {L_{st} \sum\limits_{\tau = s + 1}^{p - i} {\sum\limits_{\rho = t - 1}^{q - j} {\sum\limits_{g = s + 1}^{\tau } {\sum\limits_{h = t - 1}^{\rho } {gc_{\tau ,g} c_{\rho ,h} d_{g - 1,s} d_{h + 1,t} A(p,q,i,j,\tau ,\rho )} } } } } } \\ \, & = \mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0 < i + j < p + q} \left[ {\sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {\sum\limits_{g = 0}^{s} {\sum\limits_{h = 0}^{t} {\sum\limits_{\tau = 0}^{g - 1} {\sum\limits_{\rho = 0}^{h + 1} {gc_{s,g} c_{t,h} d_{g - 1,\tau } d_{h + 1,\rho } L_{\tau \rho } A(p - 1,q + 1,i,j,s,t)} } } } } } } \right. \\ & \left. { - \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {L_{st} \sum\limits_{\tau = s + 1}^{p - i} {\sum\limits_{\rho = t - 1}^{q - j} {\sum\limits_{g = s + 1}^{\tau } {\sum\limits_{h = t - 1}^{\rho } {gc_{\tau ,g} c_{\rho ,h} d_{g - 1,s} d_{h + 1,t} A(p,q,i,j,\tau ,\rho )} } } } } } } \right] \\ & = \mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0 < i + j < p + q} [D_{1} - D_{2} ] \\ \end{aligned} $$
(42)
where
$$ D_{1} = \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {\sum\limits_{g = 0}^{s} {\sum\limits_{h = 0}^{t} {\sum\limits_{\tau = 0}^{g - 1} {\sum\limits_{\rho = 0}^{h + 1} {gc_{s,g} c_{t,h} d_{g - 1,\tau } d_{h + 1,\rho } L_{\tau \rho } A(p - 1,q + 1,i,j,s,t)} } } } } } $$
(43)
$$ D_{2} = \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {L_{st} \sum\limits_{\tau = s + 1}^{p - i} {\sum\limits_{\rho = t - 1}^{q - j} {\sum\limits_{g = s + 1}^{\tau } {\sum\limits_{h = t - 1}^{\rho } {gc_{\tau ,g} c_{\rho ,h} d_{g - 1,s} d_{h + 1,t} A(p,q,i,j,\tau ,\rho )} } } } } } $$
(44)
By changing the order of summation and shifting the indices in D
1, we obtain
$$ D_{1} = \sum\limits_{\tau = 0}^{p - i - 2} {\sum\limits_{\rho = 0}^{q - j + 2} {L_{\tau \rho } \sum\limits_{s = \tau + 1}^{p - i - 1} {\sum\limits_{t = \rho - 1}^{q - j + 1} {\sum\limits_{g = \tau + 1}^{s} {\sum\limits_{h = \rho - 1}^{t} {gc_{s,g} c_{t,h} d_{g - 1,\tau } d_{h + 1,\rho } A(p - 1,q + 1,i,j,s,t)} } } } } } $$
(45)
Let p = p-1 and q = q + 1, (45) can be rewritten as
$$ D_{1} = \sum\limits_{\tau = 0}^{p - i - 1} {\sum\limits_{\rho = 0}^{q - j + 1} {L_{\tau \rho } \sum\limits_{s = \tau + 1}^{p - i} {\sum\limits_{t = \rho - 1}^{q - j} {\sum\limits_{g = \tau + 1}^{s} {\sum\limits_{h = \rho - 1}^{t} {gc_{s,g} c_{t,h} d_{g - 1,\tau } d_{h + 1,\rho } A(p,q,i,j,s,t)} } } } } } $$
(46)
Let s = τ, t = ρ and τ = s, ρ = t in (46), we have
$$ D_{1} = \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {L_{st} \sum\limits_{\tau = s + 1}^{p - i} {\sum\limits_{\rho = t - 1}^{q - j} {\sum\limits_{g = s + 1}^{\tau } {\sum\limits_{h = t - 1}^{\rho } {gc_{\tau ,g} c_{\rho ,h} d_{g - 1,s} d_{h + 1,t} A(p,q,i,j,\tau ,\rho )} } } } } } $$
(47)
Obviously, from (47) and (44), we can see that D
1 is equal to D
2, which means A′ in (41) can be deduced from A in (40).
Thus, changing the order of summation and shifting the indices in (39), we obtain
$$ \begin{aligned} \frac{{\partial \hat{I}_{pq} }}{\partial \beta } - \frac{{\partial I_{pq} }}{\partial \beta } &= \frac{1}{{2L_{00} }}\mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0 <i + j< p + q} \sum\limits_{s = 0}^{p - i - 1} \sum\limits_{t = 0}^{q - j + 1} L_{st} \\& \quad \times\left[ {\sum\limits_{\tau = s + 1}^{p - i} {\sum\limits_{\rho = t - 1}^{q - j} {\sum\limits_{g = s + 1}^{\tau } {\sum\limits_{h = t - 1}^{\rho } {gc_{\tau ,g} c_{\rho ,h} d_{g - 1,s} d_{h + 1,t} } } } } } \right. \\ & \quad \cdot A(p,q,i,j,\tau ,\rho ) + \sum\limits_{w = i + 1}^{p - s} {\sum\limits_{b = j - 1}^{q - t} {\sum\limits_{\lambda = i + 1}^{w} {\sum\limits_{\sigma = j - 1}^{b} {\lambda c_{w,\lambda } c_{b,\sigma } d_{\lambda - 1,i} d_{\sigma + 1,j} } } } } \cdot A(p,q,w,b,s,t) \\ & \quad \left. { - \sum\limits_{m = s + i + 1}^{p} {\sum\limits_{n = t + j - 1}^{q} {\sum\limits_{e = s + i}^{m - 1} {\sum\limits_{r = t + j}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,e} d_{n + 1,r} A(e,r,i,j,s,t)} } } } } \right] \\ &\quad = \frac{1}{{2L_{00} }}\mathop {\sum\limits_{i = 0}^{p - 1} {\sum\limits_{j = 0}^{q + 1} {I_{ij} } } }\limits_{0< i + j <p + q} \sum\limits_{s = 0}^{p - i - 1} {\sum\limits_{t = 0}^{q - j + 1} {L_{st} [W_{1} + W_{2} - W_{3} ]} } \\ \end{aligned} $$
(48)
where
$$ W_{1} = \sum\limits_{\tau = s + 1}^{p - i} {\sum\limits_{\rho = t - 1}^{q - j} {\sum\limits_{g = s + 1}^{\tau } {\sum\limits_{h = t - 1}^{\rho } {gc_{\tau ,g} c_{\rho ,h} d_{g - 1,s} d_{h + 1,t} A\left( {p,q,i,j,\tau ,\rho } \right)} } } } $$
(49)
$$ W_{2} = \sum\limits_{w = i + 1}^{p - s} {\sum\limits_{b = j - 1}^{q - t} {\sum\limits_{\lambda = i + 1}^{w} {\sum\limits_{\sigma = j - 1}^{b} {\lambda c_{w,\lambda } c_{b,\sigma } d_{\lambda - 1,i} d_{\sigma + 1,j} } } } } \cdot A(p,q,w,b,s,t) $$
(50)
$$ W_{3} = \sum\limits_{m = s + i + 1}^{p} {\sum\limits_{n = t + j - 1}^{q} {\sum\limits_{e = s + i}^{m - 1} {\sum\limits_{r = t + j}^{n + 1} {mc_{p,m} c_{q,n} d_{m - 1,e} d_{n + 1,r} A(e,r,i,j,s,t)} } } } $$
(51)
By changing the order of the summation in (49) and using the orthogonal property in (7), W
1 can be rewritten as
$$ \begin{aligned} W_{1} & = \sum\limits_{g = s + 1}^{p - i} {\sum\limits_{h = t - 1}^{q - j} {\sum\limits_{l = j}^{q - h} {\sum\limits_{n = l + h}^{q} {\sum\limits_{k = i}^{p - g} {\sum\limits_{m = k + g}^{p} {gd_{g - 1,s} d_{h + 1,t} \left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ l \hfill \\ \end{gathered} \right)c_{p,m} c_{q,n} d_{k,i} d_{l,j} \sum\limits_{\tau = g}^{m - k} {d_{m - k,\tau } c_{\tau ,g} \sum\limits_{\rho = h}^{n - l} {d_{n - l,\rho } c_{\rho ,h} } } } } } } } } \\ & = \sum\limits_{g = s + 1}^{p - i} {\sum\limits_{h = t - 1}^{q - j} {\sum\limits_{l = j}^{q - h} {\sum\limits_{n = l + h}^{q} {\sum\limits_{k = i}^{p - g} {\sum\limits_{m = k + g}^{p} {gd_{g - 1,s} d_{h + 1,t} \left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ l \hfill \\ \end{gathered} \right)c_{p,m} c_{q,n} d_{k,i} d_{l,j} \delta_{m - k,g} \delta_{n - l,h} } } } } } } \\ & = \sum\limits_{g = s + 1}^{p - i} {\sum\limits_{h = t - 1}^{q - j} {\sum\limits_{k = i}^{p - g} {\sum\limits_{l = j}^{q - h} {gd_{g - 1,s} d_{h + 1,t} \left( \begin{gathered} k + g \\ k \\ \end{gathered} \right)\left( \begin{gathered} h + l \\ l \\ \end{gathered} \right)c_{p,k + g} c_{q,h + l} d_{k,i} d_{l,j} } } } } \\ \end{aligned} $$
(52)
Let k + g = m, l + h = n and change the order of summation again, we have
$$ \begin{aligned} W_{1} & = \sum\limits_{g = s + 1}^{p - i} \sum\limits_{h = t - 1}^{q - j} \sum\limits_{m = g + i}^{p} \sum\limits_{n = h + j}^{q} gd_{g - 1,s} d_{h + 1,t} \left( \begin{gathered} m \\ m - g \\ \end{gathered} \right)\left( \begin{gathered} n \\ n - h \\ \end{gathered} \right)\\& \quad \times c_{p,m} c_{q,n} d_{m - g,i} d_{n - h,j} \\ & = \sum\limits_{m = s + i + 1}^{p} \sum\limits_{n = t + j - 1}^{q} \sum\limits_{g = s + 1}^{m - i} \sum\limits_{h = t - 1}^{n - j} gd_{g - 1,s} d_{h + 1,t} \left( \begin{gathered} m \\ m - g \\ \end{gathered} \right)\\& \quad \times \left( \begin{gathered} n \\ n - h \\ \end{gathered} \right)c_{p,m} c_{q,n} d_{m - g,i} d_{n - h,j} \\ & = \sum\limits_{g = s + i + 1}^{p} \sum\limits_{h = t + j - 1}^{q} \sum\limits_{m = s + 1}^{g - i} \sum\limits_{n = t - 1}^{h - j} md_{m - 1,s} d_{n + 1,t} \left( \begin{gathered} g \\ g - m \\ \end{gathered} \right)\\&\quad \times \left( \begin{gathered} h \\ h - n \\ \end{gathered} \right)c_{p,g} c_{q,h} d_{g - m,i} d_{h - n,j} \\ \end{aligned} $$
(53)
In a similar way, we will obtain
$$ W_{2} = \sum\limits_{g = i + s + 1}^{p} {\sum\limits_{h = j + t - 1}^{q} {\sum\limits_{m = i + 1}^{g - s} {\sum\limits_{n = j - 1}^{h - t} {md_{m - 1,i} d_{n + 1,j} \left( \begin{gathered} g \hfill \\ m \hfill \\ \end{gathered} \right)\left( \begin{gathered} h \hfill \\ n \hfill \\ \end{gathered} \right)c_{p,g} c_{q,h} d_{g - m,s} d_{h - n,t} } } } } $$
(54)
$$ W_{3} = \sum\limits_{g = i + s + 1}^{p} {\sum\limits_{h = j + t - 1}^{q} {\sum\limits_{m = i + 1}^{g - s} {\sum\limits_{n = j - 1}^{h - t} {gc_{p,g} c_{q,h} \left( \begin{gathered} g - 1 \hfill \\ m - 1 \hfill \\ \end{gathered} \right)\left( \begin{gathered} h + 1 \hfill \\ n + 1 \hfill \\ \end{gathered} \right)d_{m - 1,i} d_{g - m,s} d_{n + 1,j} d_{h - n,t} } } } } $$
(55)
Let m′ = g − m + 1, n′ = h – n – 1
$$ \begin{aligned} W_{3} - W_{2} & = \sum\limits_{g = s + i + 1}^{p} \sum\limits_{h = t + j - 1}^{q} c_{p,g} c_{q,h} \sum\limits_{m = i + 1}^{g - s} \sum\limits_{n = j - 1}^{h - t} \left[ g\left( \begin{gathered} g - 1 \\ m - 1 \\ \end{gathered} \right)\right.\\&\quad \times \left. \left( \begin{gathered} h + 1 \\ n + 1 \\ \end{gathered} \right) - m\left( \begin{gathered} g \\ m \\ \end{gathered} \right)\left( \begin{gathered} h \\ n \\ \end{gathered} \right) \right] \cdot d_{m - 1,i} d_{g - m,s} d_{n + 1,j} \\ & \quad \times d_{h - n,t} \\ & = \sum\limits_{g = s + i + 1}^{p} \sum\limits_{h = t + j - 1}^{q} c_{p,g} c_{q,h} \sum\limits_{m = i + 1}^{g - s} \sum\limits_{n = j - 1}^{h - t} m\left( \begin{gathered} g \\ m \\ \end{gathered} \right)\\&\quad\times\left( \begin{gathered} h \\ n + 1 \\ \end{gathered} \right)d_{m - 1,i} d_{g - m,s} d_{n + 1,j} d_{h - n,t} \\ & = \sum\limits_{g = s + i + 1}^{p} \sum\limits_{h = t + j - 1}^{q} c_{p,g} c_{q,h} \sum\limits_{{m^{\prime} = s + 1}}^{g - i} \sum\limits_{{n^{\prime} = t - 1}}^{h - j} (g - m^{\prime } + 1) \\ & \quad \times \left( \begin{gathered} g \\ g - m^{\prime } + 1 \\ \end{gathered} \right)\left( \begin{gathered} h \\ h - n^{\prime } \\ \end{gathered} \right) \cdot d_{{g - m^{\prime},i}} d_{{m^{\prime} - 1,s}} d_{{h - n^{\prime},j}} d_{{n^{\prime} + 1,t}} \\ & = \sum\limits_{g = s + i + 1}^{p} \sum\limits_{h = t + j - 1}^{q} c_{p,g} c_{q,h} \sum\limits_{m = s + 1}^{g - i} \sum\limits_{n = t - 1}^{h - j} (g - m + 1)\\&\quad \times \left( \begin{gathered} g \\ g - m + 1 \\ \end{gathered} \right)\left( \begin{gathered} h \\ h - n \\ \end{gathered} \right) \cdot d_{g - m,i} d_{m - 1,s} d_{h - n,j} d_{n + 1,t} \\ & = \sum\limits_{g = s + i + 1}^{p} \sum\limits_{h = t + j - 1}^{q} c_{p,g} c_{q,h} \sum\limits_{m = s + 1}^{g - i} \sum\limits_{n = t - 1}^{h - j} m\left( \begin{gathered} g \\ m \\ \end{gathered} \right)\left( \begin{gathered} h \\ n \\ \end{gathered} \right)\\&\quad\times d_{g - m,i} d_{m - 1,s} d_{h - n,j} d_{n + 1,t} \\ \end{aligned} $$
(56)
Then we can obtain W
3 − W
2 − W
1 = 0, which means \( \frac{{\partial I_{pq} }}{\partial \beta } - \frac{{\partial \hat{I}_{pq} }}{\partial \beta } = 0 \). Now the proof for X-shearing is completed.
1.2 B. Y-shearing
Because of symmetry, the proof for Y-shearing can be derived using the method similar to X-shearing.
1.3 C. Anisotropy scaling normalization
As we can see from (15), anisotropy scaling can be subdivided into two more simple steps
1. u = αx, v = y
2. u = x, v = δy
Obviously, validating (27) for each of these two steps is sufficient to the proof of the anisotropy scaling.
For step 1, we have
$$ \frac{{\partial L_{pq} }}{\partial \alpha } = \frac{\partial }{\partial \alpha }\int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {P_{p} (u)P_{q} (v)\ddot{f}(u,v){\text{d}}u{\text{d}}v} } = \frac{\partial }{\partial \alpha }\int\limits_{ - 1}^{1} {\int\limits_{ - 1}^{1} {P_{p} (\alpha x)P_{q} (y)f(x,y)\alpha {\text{d}}x{\text{d}}y} } $$
(57)
Using the similar deduction in (29) and (30), we can obtain
$$ \frac{{\partial L_{pq} }}{\partial \alpha } = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{i = 0}^{m} {\sum\limits_{j = 0}^{n} {\frac{(m + 1)}{\alpha }c_{p,m} c_{q,n} d_{m,i} d_{n,j} L_{ij} } } } } $$
(58)
Now, we need to prove that:
$$ \frac{{\partial I_{pq} }}{\partial \alpha } = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{i = 0}^{m} {\sum\limits_{j = 0}^{n} {\frac{(m + 1)}{\alpha }c_{p,m} c_{q,n} d_{m,i} d_{n,j} I_{ij} } } } } $$
(59)
Since I
pq
is equal to L
pq
, it can be easily verified that (59) is true for p + q = 1. For p + q = 3, we just consider p = 2, q = 1 in which
$$ \begin{aligned} \frac{{\partial I_{21} }}{\partial \alpha } & = \frac{{\partial L_{21} }}{\partial \alpha } - \mathop {\sum\limits_{i = 0}^{2} {\sum\limits_{j = 0}^{1} {} } }\limits_{0 < i + j < 3} \sum\limits_{s = 0}^{2 - i} {\sum\limits_{t = 0}^{1 - j} {\left[ {\frac{{\partial \left( {\frac{{L_{s,t} }}{{2L_{00} }}} \right)}}{\partial \alpha }I_{ij} + \frac{{L_{s,t} }}{{2L_{00} }}\frac{{\partial I_{ij} }}{\partial \alpha }} \right] \cdot } A(2,1,i,j,s,t)} \\ & = \sum\limits_{m = 0}^{2} {\sum\limits_{n = 0}^{1} {\sum\limits_{i = 0}^{m} {\sum\limits_{j = 0}^{n} {\frac{(m + 1)}{\alpha }c_{2,m} c_{1,n} d_{m,i} d_{n,j} I_{ij} } } } } \\ & - \mathop {\sum\limits_{i = 0}^{2} {\sum\limits_{j = 0}^{1} {} } }\limits_{0 < i + j < 3} \sum\limits_{s = 0}^{2 - i} {\sum\limits_{t = 0}^{1 - j} {\left[ {\frac{{\partial \left( {\frac{{L_{s,t} }}{{2L_{00} }}} \right)}}{\partial \alpha }I_{ij} + \frac{{L_{s,t} }}{{2L_{00} }}\frac{{\partial I_{ij} }}{\partial \alpha }} \right] \cdot } A(2,1,i,j,s,t)} \\ \end{aligned} $$
(60)
It is sufficient to prove
$$ \frac{{\partial I_{21} }}{\partial \alpha } - \sum\limits_{m = 0}^{2} {\sum\limits_{n = 0}^{1} {\sum\limits_{i = 0}^{m} {\sum\limits_{j = 0}^{n} {\frac{(m + 1)}{\alpha }c_{2,m} c_{1,n} d_{m,i} d_{n,j} I_{ij} } } } } = 0 $$
(61)
Based on Theorem 1 and (58), we have
$$ \begin{aligned} \frac{{\partial L_{00} }}{\partial \alpha } & = \frac{1}{\alpha }c_{00} c_{00} d_{00} d_{00} L_{00} \\ \frac{{\partial L_{10} }}{\partial \alpha } & = \frac{1}{\alpha }c_{10} c_{00} d_{00} d_{00} L_{00} + \frac{2}{\alpha }c_{11} c_{00} d_{10} d_{00} L_{00} + \frac{2}{\alpha }c_{11} c_{00} d_{11} d_{00} L_{10} \\ \frac{{\partial L_{01} }}{\partial \alpha } & = \frac{1}{\alpha }c_{00} c_{10} d_{00} d_{00} L_{00} + \frac{1}{\alpha }c_{00} c_{11} d_{00} d_{10} L_{00} + \frac{1}{\alpha }c_{00} c_{11} d_{11} d_{00} L_{01} \\ L_{00} & = I_{00} \quad L_{01} = I_{01} \quad L_{10} = I_{10} \\ I_{21} & = L_{21} - \frac{1}{{2L_{00} }}\{ I_{01} [L_{00} A(2,1,0,1,0,0) + L_{10} A(2,1,0,1,1,0) + L_{20} A(2,1,0,1,2,0)] \\ & + I_{10} [L_{00} A(2,1,1,0,0,0) + L_{01} A(2,1,1,0,0,1) + L_{10} A(2,1,1,0,1,0) + L_{11} A(2,1,1,0,1,1)\} \\ \end{aligned} $$
(62)
Substituting (60) and (62) into (61), we can conduct that (59) is true for p = 2, q = 1.
Assume that (31) holds for all invariants of order less than p + q, next we will consider the case of p + q.
Based on the definition of blur invariants, we have
$$ \begin{aligned} \frac{{\partial I_{pq} }}{\partial \alpha } & = \frac{{\partial L_{p,q} }}{\partial \alpha } - \mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\left[ {\frac{{\partial \left( {\frac{{L_{s,t} }}{{2L_{00} }}} \right)}}{\partial \alpha }I_{ij} + \frac{{L_{s,t} }}{{2L_{00} }}\frac{{\partial I_{ij} }}{\partial \alpha }} \right] \cdot } A(p,q,i,j,s,t)} \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{e = 0}^{m} {\sum\limits_{r = 0}^{n} {\frac{(m + 1)}{\alpha }c_{p,m} c_{q,n} d_{m,e} d_{n,r} L_{er} } } } } - \mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\left[ {\frac{{\frac{{\partial L_{s,t} }}{\partial \alpha }L_{00} - \frac{{\partial L_{0,0} }}{\partial \alpha }L_{s,t} }}{{2L_{00}^{2} }}I_{ij} } \right.} } \\ & + \frac{{L_{s,t} }}{{2L_{00} }}\sum\limits_{\lambda = 0}^{i} {\sum\limits_{\sigma = 0}^{j} {\sum\limits_{w = 0}^{\lambda } {\sum\limits_{b = 0}^{\sigma } {\left. {\frac{(\lambda + 1)}{\alpha }c_{i,\lambda } c_{j,\sigma } d_{\lambda ,w} d_{\sigma ,b} I_{wb} } \right] \cdot } } } } A(p,q,i,j,s,t) \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{e = 0}^{m} {\sum\limits_{r = 0}^{n} {\frac{(m + 1)}{\alpha }c_{p,m} c_{q,n} d_{m,e} d_{n,r} L_{er} } } } } \\ & - \mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\left[ {\sum\limits_{g = 0}^{s} {\sum\limits_{h = 0}^{t} {\sum\limits_{\tau = 0}^{g} {\sum\limits_{\rho = 0}^{h} {\frac{{(g + 1)c_{s,g} c_{t,h} d_{g,\tau } d_{h,\rho } L_{\tau \rho } }}{{2\alpha L_{00} }}I_{ij} } } } } - \frac{{L_{s,t} }}{{2\alpha L_{00} }}I_{ij} } \right.} } \\ & + \frac{{L_{s,t} }}{{2L_{00} }}\sum\limits_{\lambda = 0}^{i} {\sum\limits_{\sigma = 0}^{j} {\sum\limits_{w = 0}^{\lambda } {\sum\limits_{b = 0}^{\sigma } {\left. {\frac{(\lambda + 1)}{\alpha }c_{i,\lambda } c_{j,\sigma } d_{\lambda ,w} d_{\sigma ,b} I_{wb} } \right] \cdot } } } } A(p,q,i,j,s,t) \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{q} {\sum\limits_{e = 0}^{m} {\sum\limits_{r = 0}^{n} {\frac{(m + 1)}{\alpha }c_{p,m} c_{q,n} d_{m,e} d_{n,r} L_{er} } } } } \\ & \mathop { - \sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \frac{{I_{ij} }}{{2\alpha L_{00} }}\left[ {\sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {L_{st} \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{g = s}^{\tau } {\sum\limits_{h = t}^{\rho } {(g + 1)c_{\tau ,g} c_{\rho ,h} d_{g,s} d_{h,t} } } } } \cdot A(p,q,i,j,\tau ,\rho )} } } \right. \\ & - \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {L_{st} \cdot A(p,q,i,j,s,t)} } \\ & \left. { + \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\sum\limits_{w = i}^{p - s} {\sum\limits_{b = j}^{q - t} {\sum\limits_{\lambda = i}^{w} {\sum\limits_{\sigma = j}^{b} {(\lambda + 1)c_{w,\lambda } c_{b,\sigma } d_{\lambda ,i} d_{\sigma ,j} L_{st} \cdot } } } } A(p,q,w,b,s,t)} } } \right] \\ \end{aligned} $$
(63)
Let
$$ \begin{aligned} \frac{{\partial \hat{I}_{pq} }}{\partial \alpha } & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{p} {\sum\limits_{e = 0}^{m} {\sum\limits_{r = 0}^{n} {\frac{(m + 1)}{\alpha }c_{pm} c_{qn} d_{me} d_{nr} I_{er} } } } } \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{p} {\sum\limits_{e = 0}^{m} {\sum\limits_{r = 0}^{n} {\frac{(m + 1)}{\alpha }c_{pm} c_{qn} d_{me} d_{nr} } } } } \\&\quad\times \left[ {L_{er} - \frac{1}{{2L_{00} }}\mathop{\sum\limits_{i = 0}^{e} {\sum\limits_{j = 0}^{r} {} } }\limits_{0 < i + j < e + q} rI_{ij} \sum\limits_{s = 0}^{e - i} {\sum\limits_{t = 0}^{r - j} {L_{s,t} } } } { \cdot A(e,r,i,j,s,t)} \right] \\ & = \sum\limits_{m = 0}^{p} {\sum\limits_{n = 0}^{p} {\sum\limits_{e = 0}^{m} {\sum\limits_{r = 0}^{n} {\frac{(m + 1)}{\alpha }c_{pm} c_{qn} d_{me} d_{nr} L_{er} } } } } \\ & \quad - \frac{1}{{2\alpha L_{00} }}\mathop{\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} I_{ij} \sum\limits_{s = 0}^{p - i} \sum\limits_{t = 0}^{q - j} L_{st} \sum\limits_{m = s + i}^{p} \sum\limits_{n = t + j}^{q} \sum\limits_{e = s + i}^{m} \sum\limits_{r = t + j}^{n}\\&\quad\times {(m + 1)c_{pm} c_{qn} d_{me} d_{nr} L_{st} } \cdot A(e,r,i,j,s,t) \\ \end{aligned} $$
(64)
If we can prove\( \frac{{\partial I_{pq} }}{\partial \alpha } - \frac{{\partial \hat{I}_{pq} }}{\partial \alpha } = 0 \), then (59) can be proved. Thus
$$ \begin{aligned} \frac{{\partial I_{pq} }}{\partial \alpha } - \frac{{\partial \hat{I}_{pq} }}{\partial \alpha } & = \frac{1}{{2\alpha L_{00} }}\mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} I_{ij} \sum\limits_{s = 0}^{p - i} \sum\limits_{t = 0}^{q - j} L_{st} \\&\quad\times\sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {\sum\limits_{e = s + i}^{m} {\sum\limits_{r = t + j}^{n} {(m + 1)c_{p,m} c_{q,n} d_{m,e} d_{n,r} } } } } \\ & \quad\mathop{ \cdot A(e,r,i,j,s,t) - \sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \frac{{I_{ij} }}{{2\alpha L_{00} }}\left[ {\sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {L_{st} \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{g = s}^{\tau } {\sum\limits_{h = t}^{\rho } {(g + 1)c_{\tau ,g} c_{\rho ,h} } } } } } } } \right. \cdot d_{g,s} d_{h,t} \\&\quad\times A(p,q,i,j,\tau ,\rho ) - \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {L_{st} \cdot A(p,q,i,j,s,t)} } \\ & \left. {\quad + \sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {\sum\limits_{w = i}^{p - s} {\sum\limits_{b = j}^{q - t} {\sum\limits_{\lambda = i}^{w} {\sum\limits_{\sigma = j}^{b} {(\lambda + 1)c_{w,\lambda } c_{b,\sigma } d_{\lambda ,i} d_{\sigma ,j} L_{st} \cdot } } } } A(p,q,w,b,s,t)} } } \right] \\ & = \mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \frac{{I_{ij} }}{{2\alpha L_{00} }}\sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {L_{st} \left[ {\sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {\sum\limits_{e = s + i}^{m} {\sum\limits_{r = t + j}^{n} {(m + 1)c_{pm} c_{qn} d_{me} d_{yr} } } } } } \right.} } \\ & \cdot A(e,r,i,j,s,t) - \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{g = s}^{\tau } {\sum\limits_{h = t}^{\rho } {(g + 1)c_{\tau g} c_{\rho h} d_{gs} d_{ht} } } } } \cdot A(p,q,i,j,\tau ,\rho ) - \\ & \left. {A(p,q,i,j,s,t) - \sum\limits_{w = i}^{p - s} {\sum\limits_{b = j}^{q - t} {\sum\limits_{\lambda = i}^{w} {\sum\limits_{\sigma = j}^{b} {(\lambda + 1)c_{w\lambda } c_{b\sigma } d_{\lambda i} d_{\sigma j} \cdot } } } } A(p,q,w,b,s,t)} \right] \\ & = \mathop {\sum\limits_{i = 0}^{p} {\sum\limits_{j = 0}^{q} {} } }\limits_{0 < i + j < p + q} \frac{{I_{ij} }}{{2\alpha L_{00} }}\sum\limits_{s = 0}^{p - i} {\sum\limits_{t = 0}^{q - j} {L_{st} [T_{1} - T_{2} - T_{3} ]} } \\ \end{aligned} $$
(65)
where
$$ \begin{aligned} T_{1} & = \sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {\sum\limits_{e = s + i}^{m} {\sum\limits_{r = t + j}^{n} {(m + 1)c_{pm} c_{qn} d_{me} d_{yr} } } } } \cdot A(e,r,i,j,s,t) \\ T_{2} & = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{g = s}^{\tau } {\sum\limits_{h = t}^{\rho } {(g + 1)c_{\tau g} c_{\rho h} d_{gs} d_{ht} } } } } \cdot A(p,q,i,j,\tau ,\rho ) - A(p,q,i,j,s,t) \\ T_{3} & = \sum\limits_{w = i}^{p - s} {\sum\limits_{b = j}^{q - t} {\sum\limits_{\lambda = i}^{w} {\sum\limits_{\sigma = j}^{b} {(\lambda + 1)c_{w\lambda } c_{b\sigma } d_{\lambda i} d_{\sigma j} \cdot } } } } A(p,q,w,b,s,t) \\ \end{aligned} $$
After changing the variables, we have
$$ \begin{aligned} T_{2} & = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{g = \tau }^{p - i} {\sum\limits_{h = \rho }^{q - j} {(\tau + 1)c_{g\tau } c_{h\rho } d_{\tau s} d_{\rho t} } } } } \cdot A(p,q,i,j,g,h) - A(p,q,i,j,s,t) \\ & = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{g = \tau }^{p - i} {\sum\limits_{h = \rho }^{q - j} {(\tau + 1)c_{g\tau } c_{h\rho } d_{\tau s} d_{\rho t} } } } } \cdot \sum\limits_{k = i}^{p - g} {\sum\limits_{m = k + g}^{p} {\sum\limits_{l = j}^{q - h} {\sum\limits_{n = l + h}^{q} {\left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ l \hfill \\ \end{gathered} \right)c_{pm} c_{qn} d_{ki} } } } } d_{m - k,g} d_{lj} d_{n - l,h} \\ & \quad - A(p,q,i,j,s,t) \\ \end{aligned} $$
(66)
By changing the order of the summation in (66) and using the orthogonal property in (7), T
2 can be rewritten as
$$ \begin{gathered} T_{2} = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{k = i}^{p - \tau } {\sum\limits_{l = j}^{q - \rho } {(\tau + 1)d_{\tau s} d_{\rho t} } } } } \left( \begin{gathered} k + \tau \\ k \\ \end{gathered} \right)\left( \begin{gathered} \rho + l \\ l \\ \end{gathered} \right)c_{p,k + \tau } c_{q,\rho + l} d_{ki} d_{lj} \\ - \sum\limits_{k = i}^{p - s} {\sum\limits_{l = j}^{q - t} {\sum\limits_{m = k + s}^{p} {\sum\limits_{n = l + t}^{q} {\left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ l \hfill \\ \end{gathered} \right)c_{pm} c_{qn} d_{ki} } } } } d_{m - k,s} d_{lj} d_{n - l,t} \\ = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{k = i}^{p - \tau } {\sum\limits_{l = j}^{q - \rho } {(\tau + 1)d_{\tau s} d_{\rho t} } } } } \left( \begin{gathered} k + \tau \\ k \\ \end{gathered} \right)\left( \begin{gathered} \rho + l \\ l \\ \end{gathered} \right)c_{p,k + \tau } c_{q,\rho + l} d_{ki} d_{lj} \\ - \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{m = \tau + i}^{p} {\sum\limits_{n = \rho + j}^{q} {\left( \begin{gathered} m \hfill \\ \tau \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ \rho \hfill \\ \end{gathered} \right)c_{pm} c_{qn} d_{\tau s} } } } } d_{m - \tau ,i} d_{\rho t} d_{n - \rho ,j} \\ \end{gathered} $$
(67)
Let m = k+τ and n = l+ρ
$$ \begin{aligned} T_{2} & = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{k = i}^{p - \tau } {\sum\limits_{l = j}^{q - \rho } {(\tau + 1)d_{\tau s} d_{\rho t} } } } } \left( \begin{gathered} k + \tau \\ k \\ \end{gathered} \right)\left( \begin{gathered} \rho + l \\ l \\ \end{gathered} \right)c_{p,k + \tau } c_{q,\rho + l} d_{ki} d_{lj} \\ & \quad - \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{k = i}^{p - \tau } {\sum\limits_{l = j}^{q - \rho } {\left( \begin{gathered} k + \tau \\ \tau \\ \end{gathered} \right)\left( \begin{gathered} l + \rho \\ \rho \\ \end{gathered} \right)c_{p,k + \tau } c_{q,l + \rho } d_{\tau s} } } } } d_{ki} d_{\rho t} d_{lj} \\ & = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{k = i}^{p - \tau } {\sum\limits_{l = j}^{q - \rho } {\tau d_{\tau s} d_{\rho t} } } } } \left( \begin{gathered} k + \tau \\ k \\ \end{gathered} \right)\left( \begin{gathered} \rho + l \\ l \\ \end{gathered} \right)c_{p,k + \tau } c_{q,\rho + l} d_{ki} d_{lj} \\ & = \sum\limits_{\tau = s}^{p - i} {\sum\limits_{\rho = t}^{q - j} {\sum\limits_{m = i + \tau }^{p} {\sum\limits_{n = j + \rho }^{q} \tau } } } \left( \begin{gathered} m \\ \tau \\ \end{gathered} \right)\left( \begin{gathered} n \\ n - \rho \\ \end{gathered} \right)c_{p,m} c_{q,n} d_{\tau s} d_{\rho t} d_{m - \tau ,i} d_{n - \rho ,j} \\ & = \sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {\sum\limits_{\tau = s}^{m - i} {\sum\limits_{\rho = t}^{n - j} \tau } } } \left( \begin{gathered} m \\ \tau \\ \end{gathered} \right)\left( \begin{gathered} n \\ n - \rho \\ \end{gathered} \right)c_{p,m} c_{q,n} d_{\tau s} d_{\rho t} d_{m - \tau ,i} d_{n - \rho ,j} \\ & = \sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {\sum\limits_{k = s}^{m - i} {\sum\limits_{l = t}^{n - j} k } } } \left( \begin{gathered} m \\ k \\ \end{gathered} \right)\left( \begin{gathered} n \\ n - l \\ \end{gathered} \right)c_{p,m} c_{q,n} d_{m - k,s} d_{n - l,t} d_{k,i} d_{l,j} \\ \end{aligned} $$
(68)
Similarly
$$ T3 = \sum\limits_{m = i + s}^{p} {\sum\limits_{n = t + j}^{q} {\sum\limits_{k = i}^{m - s} {\sum\limits_{l = j}^{n - t} {(k + 1)} } } } \left( \begin{gathered} m \\ k \\ \end{gathered} \right)\left( \begin{gathered} n \\ l \\ \end{gathered} \right)c_{p,m} c_{q,n} d_{m - k,s} d_{n - l,t} d_{ki} d_{lj} $$
(69)
$$ T1 = \sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {\sum\limits_{k = i}^{m - s} {\sum\limits_{l = j}^{n - t} {(m + 1)\left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ l \hfill \\ \end{gathered} \right)c_{pm} c_{qn} d_{m - k,s} d_{n - l,t} d_{ki} d_{lj} } } } } $$
(70)
Therefore
$$ \begin{aligned} T1 - T2 - T3 & = \sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {c_{pm} c_{qn} } } \left[ {\sum\limits_{k = i}^{m - s} {\sum\limits_{l = j}^{n - t} {(m - k)\left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ l \hfill \\ \end{gathered} \right)d_{m - k,s} d_{n - l,t} d_{ki} d_{lj} } } } \right. \\ & \left. {\quad - \sum\limits_{k = s}^{m - i} {\sum\limits_{l = t}^{n - j} {k\left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \\ n - l \\ \end{gathered} \right)d_{m - k,s} d_{n - l,t} d_{ki} d_{lj} } } } \right] \\ \end{aligned} $$
(71)
Let x-k = k′, y-l = l′
$$ \begin{aligned} T1 - T2 - T3 & = \sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {c_{pm} c_{qn} } } \left[ \sum\limits_{{k^{\prime } = s}}^{m - i} \sum\limits_{{l^{\prime } = t}}^{n - j} k^{\prime } \left( \begin{gathered} m \\ m - k^{\prime } \\ \end{gathered} \right) \right. \\&\quad\times \left( \begin{gathered} n \\ n - l^{\prime } \\ \end{gathered} \right)d_{{k^{\prime } ,s}} d_{{l^{\prime } ,t}} d_{{m - k^{\prime } ,i}} d_{{n - l^{\prime } ,j}} \\ &\quad \left.{- \sum\limits_{k = s}^{m - i} {\sum\limits_{l = t}^{n - j} {k\left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \\ n - l \\ \end{gathered} \right)d_{m - k,s} d_{n - l,t} d_{ki} d_{lj} } } } \right] \\ & = \sum\limits_{m = s + i}^{p} {\sum\limits_{n = t + j}^{q} {c_{pm} c_{qn} } } \sum\limits_{k = s}^{m - i} \sum\limits_{l = t}^{n - j} k\left( \begin{gathered} m \hfill \\ k \hfill \\ \end{gathered} \right)\left( \begin{gathered} n \hfill \\ l \hfill \\ \end{gathered} \right)\\&\quad\times \left[ {d_{k,s} d_{l,t} d_{m - k,i} d_{n - l,j} - d_{m - k,s} d_{n - l,t} d_{ki} d_{lj} } \right] = 0 \\ \end{aligned} $$
(72)
Then we can obtain \( \frac{{\partial I_{pq} }}{\partial \alpha } - \frac{{\partial \hat{I}_{pq} }}{\partial \alpha } = 0 \). Now the proof for step 1 is completed.
Because of the symmetry, (27) is true for step 2 which can be proved using the similar method as step 1. Finally, we can conclude that (27) holds for anisotropy scaling.
Now, the proof for Theorem 1 has been completed.
