The Shannon Total Variation

Abstract

Discretization schemes commonly used for total variation regularization lead to images that are difficult to interpolate, which is a real issue for applications requiring subpixel accuracy and aliasing control. In the present work, we reconciliate total variation with Shannon interpolation and study a Fourier-based estimate that behaves much better in terms of grid invariance, isotropy, artifact removal and subpixel accuracy. We show that this new variant (called Shannon total variation) can be easily handled with classical primal–dual formulations and illustrate its efficiency on several image processing tasks, including deblurring, spectrum extrapolation and a new aliasing reduction algorithm.

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Acknowledgements

We would like to thank the anonymous reviewers for their valuable suggestions.

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Correspondence to Rémy Abergel.

Appendices

Appendix 1: Proof of Proposition 1

Let us consider, for \(x\in {\mathbb R}\setminus {\mathbb Z}\),

$$\begin{aligned} S_n(x)= & {} \sum _{p=-n}^{n} \mathrm {sinc}(x-pM)\\= & {} \mathrm {sinc}(x) + \sum _{p=1}^{n} \mathrm {sinc}(x-pM) + \mathrm {sinc}(x+pM)\\= & {} \frac{\sin \pi x}{\pi x} + \sum _{p=1}^{n} (-1)^{pM} \left( \frac{\sin \pi x}{\pi (x-pM)} + \frac{\sin \pi x}{\pi (x+pM)} \right) . \end{aligned}$$

Writing \(x= \frac{M t}{\pi }\), we obtain

$$\begin{aligned} S_n(x) = \frac{\sin Mt}{M} \left( \frac{1}{t} + \sum _{p=1}^{n} (-1)^{pM} \left( \frac{1}{t-p\pi } + \frac{1}{t+p \pi } \right) \right) \end{aligned}$$

and the limit \(\mathrm {sincd}_M(x) = \lim _{n\rightarrow \infty } S_n(x)\) can be computed explicitly using classical series expansions (due to Euler):

$$\begin{aligned} \forall t\in {\mathbb R}\setminus \pi {\mathbb Z}, \;\;\; \displaystyle \frac{1}{\tan t}= & {} \frac{1}{t} + \sum _{p=1}^\infty \frac{1}{t-p\pi } + \frac{1}{t+p\pi },\\ \displaystyle \frac{1}{\sin t}= & {} \frac{1}{t} + \sum _{p=1}^\infty (-1)^p \left( \frac{1}{t-p\pi } + \frac{1}{t+p\pi }\right) . \end{aligned}$$

If M is odd, \((-1)^{pM}=(-1)^p\) and we obtain

$$\begin{aligned} \mathrm {sincd}_M(x) = \frac{\sin Mt}{M \sin t} = \frac{\sin \pi x}{M\sin \frac{\pi x}{M}}, \end{aligned}$$

and if M is even, \((-1)^{pM}=1\) and the other series yields

$$\begin{aligned} \mathrm {sincd}_M(x) = \frac{\sin Mt}{M \tan t} = \frac{\sin \pi x}{M\tan \frac{\pi x}{M}} \end{aligned}$$

as announced. \(\square \)

Appendix 2: Proof of Theorem 2

Since each operator \(T_z\) is linear and translation-invariant (Hypothesis (ii)), it can be written as a convolution, that is,

$$\begin{aligned} T_z s(k) = (\psi _z \star s)(k) := \sum _{l\in I_M} \psi _z(k-l) s(l), \end{aligned}$$
(67)

where \(\psi _z\) is an element of \(\mathcal S\). Taking the DFT of (67), we obtain

$$\begin{aligned} \forall \alpha \in {\mathbb Z}, \quad \widehat{T_z s}(\alpha ) = \widehat{\psi }_z(\alpha ) \widehat{s}(\alpha ). \end{aligned}$$
(68)

Now, from Hypothesis (iii) we immediately get

$$\begin{aligned} \forall z,w \in {\mathbb R},\; \forall \alpha \in {\mathbb Z},\quad \widehat{\psi }_{z+w}(\alpha ) = \widehat{\psi }_z(\alpha ) \widehat{\psi }_w(\alpha ), \end{aligned}$$
(69)

and by continuity of \(z\mapsto \widehat{\psi }_z(\alpha )\) (deduced from Hypothesis (i)) we obtain

$$\begin{aligned} \forall \alpha \in {\mathbb Z}, \quad \widehat{\psi }_z(\alpha ) = \mathrm{e}^{\gamma (\alpha ) z} \end{aligned}$$
(70)

for some \(\gamma (\alpha ) \in {\mathbb C}\). Since \(\widehat{\psi }_1(\alpha ) = \mathrm{e}^\frac{-2i\pi \alpha }{M}\), we have

$$\begin{aligned} \gamma (\alpha ) = -2i\pi \left( \frac{\alpha }{M} + p(\alpha ) \right) , \end{aligned}$$
(71)

where \(p(\alpha ) \in {\mathbb Z}\) and \(p(-\alpha )=-p(\alpha )\) (the fact that \(T_z u\) is real-valued implies that \(\widehat{\psi }_z(-\alpha )=\widehat{\psi }_z(\alpha )^*\)).

Last, we compute

$$\begin{aligned} \Vert T_z - id \Vert _2^2= & {} \sup _{\Vert s\Vert _2=1}\Vert T_z s- s \Vert _2^2 \\= & {} \frac{1}{M}\sup _{\Vert s\Vert _2=1}\Vert \widehat{T_z s}- \hat{s} \Vert _2^2\\= & {} \frac{1}{M} \sup _{\Vert \hat{s}\Vert _2^2=M} \sum _{\alpha \in \widehat{I}_M} |\mathrm{e}^{-2i\pi \left( \frac{\alpha }{M} + p(\alpha ) \right) z} -1|^2 \cdot |\hat{s}(\alpha )|^2\\= & {} 4 \max _{\alpha \in \widehat{I}_M} \sin ^2 \left( \pi \left( \frac{\alpha }{M} + p(\alpha ) \right) z \right) \\= & {} 4 \pi ^2 z^2 \max _{\alpha \in \widehat{I}_M} \left( \frac{\alpha }{M} + p(\alpha ) \right) ^2 + \underset{z\rightarrow 0}{o}(z^2). \end{aligned}$$

Hence,

$$\begin{aligned} \lim _{z\rightarrow 0} \;|z|^{-1} \Vert T_z - id \Vert _2 = 2 \pi \max _{\alpha \in \widehat{I}_M} \left| \frac{\alpha }{M} + p(\alpha ) \right| \end{aligned}$$
(72)

and since \(\frac{\alpha }{M}\in (-\frac{1}{2},\frac{1}{2})\) and \(p(\alpha )\in {\mathbb Z}\) for any \(\alpha \in \widehat{I}_M\), the right-hand term of (72) is minimal if and only if \(p(\alpha )=0\) for all \(\alpha \in \widehat{I}_M\). We conclude from (71) and (70) that

$$\begin{aligned} \forall \alpha \in \widehat{I}_M, \quad \widehat{\psi }_z(\alpha ) = \mathrm{e}^{-2i\pi \alpha z/M}, \end{aligned}$$
(73)

and thus (68) can be rewritten as

$$\begin{aligned} T_z s(k) = \frac{1}{M} \sum _{\alpha \in \widehat{I}_M} \widehat{s}(\alpha ) \mathrm{e}^{-2i\pi \alpha z/M} \mathrm{e}^{-2i\pi \alpha k/M}, \end{aligned}$$
(74)

which is exactly \(S(k-z)\) thanks to (13) (recall that the real part is not needed because M is odd). Therefore, (24) is a necessary form for a set of operators \((T_z)\) satisfying Hypotheses (i) to (iv).

Conversely, one easily checks that the operators \((T_z)\) defined by (24) satisfy the Hypotheses (i) to (iv). \(\square \)

Appendix 3: Proof of Proposition 8

Let us denote by \(\nabla _{n,x} u \) and \(\nabla _{n,y} u\) the two elements of \({\mathbb R}^{\Omega _n}\) such that \(\nabla _nu = \left( \nabla _{n,x}u, \nabla _{n,y} u \right) \). In the following, the notation \(\langle \cdot ,\cdot \rangle _X\) stands for the usual Euclidean (respectively, Hermitian) inner product over the real (respectively, complex) Hilbert space X. We have

$$\begin{aligned} \langle \nabla _nu, p \rangle _{{\mathbb R}^{\Omega _n} \times {\mathbb R}^{\Omega _n}} = \langle \nabla _{n,x} u, p_x \rangle _{{\mathbb R}^{\Omega _n}} + \langle \nabla _{n,y} u, p_y \rangle _{{\mathbb R}^{\Omega _n}}. \end{aligned}$$

Recall that we defined \(\mathrm {div}_{n} = -\nabla _n^*\), the opposite of the adjoint of \(\nabla _n\). Noting \(\mathrm {div}_{n,x} = - \nabla _{n,x}^*\) and \(\mathrm {div}_{n,y} = - \nabla _{n,y}^*\), we have

$$\begin{aligned} \langle \nabla _nu, p \rangle _{{\mathbb R}^{\Omega _n} \times {\mathbb R}^{\Omega _n}} = \langle u, -\mathrm {div}_{n,x}(p_x)-\mathrm {div}_{n,y}(p_y) \rangle _{{\mathbb R}^\Omega }. \end{aligned}$$

so that we identify \(\mathrm {div}_n(p) = \mathrm {div}_{n,x}(p_x)+\mathrm {div}_{n,y}(p_y)\). Let us focus on the computation of \(\mathrm {div}_{n,x}(p_x)\). Let \(\widehat{\Omega _1}, \widehat{\Omega _2}, \widehat{\Omega _3}, \widehat{\Omega _4}\) be the sets defined by

$$\begin{aligned} \widehat{\Omega _1}= & {} \left\{ \left( \alpha ,\beta \right) \in {\mathbb R}^2,~ |\alpha |< \frac{M}{2}, ~|\beta |< \frac{N}{2} \right\} \cap {\mathbb Z}^2 \\ \widehat{\Omega _2}= & {} \left\{ \left( \pm \frac{M}{2}, \beta \right) \in {\mathbb R}^2, ~ |\beta |< \frac{N}{2} \right\} \cap {\mathbb Z}^2 \\ \widehat{\Omega _3}= & {} \left\{ \left( \alpha ,\pm \frac{N}{2}\right) \in {\mathbb R}^2,~ |\alpha | < \frac{M}{2} \right\} \cap {\mathbb Z}^2 \\ \widehat{\Omega _4}= & {} \left\{ \left( \pm \frac{M}{2}, \pm \frac{N}{2}\right) \right\} \cap {\mathbb Z}^2. \\ \end{aligned}$$

Notice that some sets among \(\widehat{\Omega _2}, \widehat{\Omega _3}\) and \(\widehat{\Omega _4}\) may be empty according to the parity of M and N. Now, let \(h_{\widehat{p_x}}\) be the function defined in Proposition 8 and let us show that

$$\begin{aligned} \forall (\alpha ,\beta ) \in \widehat{\Omega }, \quad \widehat{\mathrm {div}_{n,x}(p_x)}(\alpha ,\beta ) = 2 i \pi \frac{\alpha }{M} h_{\widehat{p_x}}(\alpha ,\beta ). \end{aligned}$$
(75)

Given \(z \in {\mathbb C}\), we denote as usual by \(z^*\) the conjugate of z. Thanks to Parseval identity, and using Proposition 6 (because we assumed \(n\ge 2\)), we have

$$\begin{aligned}&\langle \nabla _{n,x} u, p_x \rangle _{{\mathbb R}^{\Omega _n}}\\&\quad = \dfrac{1}{n^2 M N} \langle \widehat{\nabla _{n,x} u}, \widehat{p_x} \rangle _{{\mathbb C}^{\Omega _n}} \\&\quad = \dfrac{1}{n^2 MN} \displaystyle {\sum _{(\alpha ,\beta ) \in \widehat{\Omega }_n}} \widehat{\nabla _{n,x} u}(\alpha ,\beta ) \, \left( \widehat{p_x}(\alpha ,\beta )\right) ^* \\&\quad = \dfrac{1}{MN} \displaystyle {\sum \limits _{\begin{array}{c} -\frac{M}{2} \le \alpha \le \frac{M}{2} \\ -\frac{N}{2} \le \beta \le \frac{N}{2} \end{array}}} -\widehat{u}(\alpha ,\beta ) \left( 2 i \pi \varepsilon _M(\alpha ) \varepsilon _N(\beta ) \frac{\alpha }{M} \widehat{p_x}(\alpha ,\beta ) \right) ^*. \end{aligned}$$

It follows that

$$\begin{aligned} \langle \nabla _{n,x} u, p_x \rangle _{{\mathbb R}^{\Omega _n}} = S_1 + S_2 + S_3 + S_4, \end{aligned}$$

where for all \(k \in \{1,2,3,4\}\), we have set

$$\begin{aligned} S_k = \dfrac{1}{MN} \! \displaystyle {\sum _{(\alpha ,\beta ) \in \widehat{\Omega }_k}} \! -\widehat{u}(\alpha ,\beta ) \left( 2 i \pi \varepsilon _M(\alpha ) \varepsilon _N(\beta ) \frac{\alpha }{M} \widehat{p_x}(\alpha ,\beta ) \right) ^*. \end{aligned}$$

Consider \(S_1\) first. Since we have \(\varepsilon _M(\alpha ) = \varepsilon _N(\beta ) = 1\) and \(h_{\widehat{p_x}}(\alpha ,\beta ) = \widehat{p_x}(\alpha ,\beta )\) for all \((\alpha ,\beta ) \in \widehat{\Omega _1}\), we recognize

$$\begin{aligned} S_1 = \dfrac{1}{MN} \displaystyle {\sum _{|\alpha |< \frac{M}{2},~|\beta | < \frac{N}{2}}} -\widehat{u}(\alpha ,\beta ) \left( 2i\pi \frac{\alpha }{M} h_{\widehat{p_x}}(\alpha ,\beta )\right) ^*. \end{aligned}$$

Now consider \(S_2\). If M is odd, \(\widehat{\Omega }_2\) is empty and \(S_2 = 0\). Otherwise, since \(\varepsilon _M(\alpha ) \varepsilon _N(\beta ) = 1/2\) for all \((\alpha ,\beta ) \in \widehat{\Omega _2}\), by grouping together the terms \(\left( -\frac{M}{2},\beta \right) \) and \(\left( \frac{M}{2},\beta \right) \), we get

$$\begin{aligned} S_2= & {} \dfrac{1}{MN} \displaystyle {\sum _{\alpha = -\frac{M}{2}, |\beta |< \frac{N}{2} }} -\widehat{u}(\alpha ,\beta ) \\&\times \,\left( 2 i \pi \frac{1}{2} \frac{\alpha }{M} \widehat{p_x}(\alpha ,\beta ) - 2 i \pi \frac{1}{2} \frac{\alpha }{M} \widehat{p_x}(-\alpha ,\beta ) \right) ^*\\= & {} \dfrac{1}{MN} \displaystyle {\sum _{\alpha = -\frac{M}{2}, |\beta | < \frac{N}{2} }} -\widehat{u}(\alpha ,\beta ) \left( 2i\pi \frac{\alpha }{M} h_{\widehat{p_x}}(\alpha ,\beta )\right) ^*,~~~ \end{aligned}$$

since we have set \(h_{\widehat{p_x}}(-\frac{M}{2},\beta ) = \frac{1}{2} \left( \widehat{p_x}(-\frac{M}{2},\beta ) - \widehat{p_x}(\frac{M}{2},\beta )\right) \) for \(|\beta | < N/2\).

Similarly for the term \(S_3\). When N is odd, \(\widehat{\Omega _3}= \emptyset \) and \(S_3 = 0\). Otherwise, when N is even, we have \(\varepsilon _M(\alpha )\varepsilon _N(\beta ) = 1/2\) for all \((\alpha ,\beta ) \in \widehat{\Omega _3}\), thus, by grouping together the terms \(\left( \alpha ,-\frac{N}{2}\right) \) and \(\left( \alpha ,\frac{N}{2}\right) \), we get

$$\begin{aligned} S_3= & {} \dfrac{1}{MN} \displaystyle {\sum _{|\alpha |< \frac{M}{2}, \beta =-\frac{N}{2}}} -\widehat{u}(\alpha ,\beta ) \\&\times \left( 2 i \pi \frac{1}{2} \frac{\alpha }{M} \widehat{p_x}(\alpha ,\beta ) + 2 i \pi \frac{1}{2} \frac{\alpha }{M} \widehat{p_x}(\alpha ,-\beta ) \right) ^*\\= & {} \dfrac{1}{MN} \displaystyle {\sum _{|\alpha | < \frac{M}{2}, \beta = -\frac{N}{2}}} -\widehat{u}(\alpha ,\beta ) \left( 2i\pi \frac{\alpha }{M} h_{\widehat{p_x}}(\alpha ,\beta )\right) ^*, \end{aligned}$$

since we have set \(h_{\widehat{p_x}}(\alpha ,-\frac{N}{2}) = \frac{1}{2} \left( \widehat{p_x}(\alpha ,-\frac{N}{2}) + \widehat{p_x}(\alpha ,\frac{N}{2})\right) \) for \(|\alpha | < M/2\).

Lastly, let us consider \(S_4\). When M and N are both even (otherwise \(\widehat{\Omega _4} = \emptyset \) and \(S_4=0\)), we immediately get, for \(\alpha =-\frac{M}{2}\) and \(\beta =-\frac{N}{2}\),

$$\begin{aligned} S_4= & {} - \widehat{u}(\alpha ,\beta ) \left( \sum _{s_1 = \pm 1, s2 = \pm 1} 2 i \pi \frac{1}{4} s_1 \frac{\alpha }{M} \widehat{p_x}(s_1 \alpha ,s_2 \beta ) \right) ^* \\= & {} - \widehat{u}(\alpha ,\beta ) \left( 2 i \pi \frac{\alpha }{M} h_{\widehat{p_x}}(\alpha ,\beta ) \right) ^*, \end{aligned}$$

since for all \((\alpha ,\beta ) \in \widehat{\Omega _4}\), we have \(\varepsilon _M(\alpha ) \varepsilon _N(\beta ) = 1/4\) and we have set \(h_{\widehat{p_x}}(\alpha ,\beta ) = \frac{1}{4} \sum _{s_1 = \pm 1, s2 = \pm 1} s_1 \widehat{p_x}(s_1 \alpha ,s_2 \beta )\).

Finally, we can write \(S_1 + S_2 + S_3 + S_4\) as a sum over \(\widehat{\Omega }\), indeed,

$$\begin{aligned} \langle \nabla _{n,x} u, p_x \rangle _{{\mathbb R}^\Omega }= & {} S_1 + S_2 + S_3 + S_4 \\= & {} \frac{1}{MN} \sum _{(\alpha ,\beta ) \in \widehat{\Omega }} - \widehat{u}(\alpha ,\beta ) \left( 2 i \pi \frac{\alpha }{M} h_{\widehat{p_x}}(\alpha ,\beta ) \right) ^*, \end{aligned}$$

and using again the Parseval identity, we get (75). With a similar approach, one can check that

$$\begin{aligned} \forall (\alpha ,\beta ) \in \widehat{\Omega }, \quad \widehat{\mathrm {div}_{n,y}(p_y)}(\alpha ,\beta ) = 2 i \pi \frac{\beta }{N} h_{\widehat{p_y}}(\alpha ,\beta ), \end{aligned}$$

where \(h_{\widehat{p_y}}\) is defined in Proposition 8. Consequently, for any \((\alpha ,\beta ) \in \widehat{\Omega }\), we have

$$\begin{aligned} \widehat{\mathrm {div}_n(p)}(\alpha ,\beta ) = 2 i \pi \left( \frac{\alpha }{M} h_{\widehat{p_x}}(\alpha ,\beta ) + \frac{\beta }{N} h_{\widehat{p_y}}(\alpha ,\beta ) \right) , \end{aligned}$$

which ends the proof of Proposition 8. \(\square \)

Appendix 4: Proof of Theorem 3

Recall that for any integer M, we denote by \(T_M\) the real vector space of real-valued trigonometric polynomials that can be written as complex linear combination of the family \(( x \mapsto \mathrm{e}^{2 i \pi \frac{\alpha x}{M}} )_{-\frac{M}{2} \le \alpha \le \frac{M}{2}}\). In order to prove Theorem 3, we need the following Lemma.

Lemma 1

Let \(M=2m+1\) be an odd positive integer. The functions F and G defined by,

$$\begin{aligned} \forall x \in {\mathbb R}, \quad F(x) = \frac{1}{M} \sum _{\alpha =-m}^m \! \mathrm{e}^{\frac{2i\pi \alpha x}{M}}, \quad G(x) = F(x)-F(x-1), \end{aligned}$$

are both in \(T_M\) and G satisfies

$$\begin{aligned} \sum _{k=0}^{M-1} |G(k)| = 2, \quad \int _1^{M} |G(x)|\,\mathrm{d}x \ge \frac{8}{\pi ^2} \log \left( \frac{2M}{\pi }\right) -2. \end{aligned}$$

Proof

F is in \(T_M\) by construction, and so is G as the difference of two elements of \(T_M\). Writing \(\omega = \frac{\pi }{M}\), we can notice that \(F(0)=1\) and

$$\begin{aligned} \forall x \in (0,M), \quad F(x) = \frac{\mathrm{e}^{2 i \omega (-m) x}}{M} \cdot \frac{1-\mathrm{e}^{2i\pi x}}{1-\mathrm{e}^{2i\omega x}} = \frac{\sin {(\pi x)}}{M \sin {(\omega x)}}, \end{aligned}$$

so that \(F(k) = 0\) for all integers \(k \in [1,M-1]\). Consequently, \(G(0)=1, G(1)=-1\) and \(G(k)=0\) for all integers \(k \in [2,M-1]\), thus

$$\begin{aligned} \sum _{k=0}^{M-1} |G(k)| = |G(0)| + |G(1)| = 2, \end{aligned}$$

yielding the first announced result of the Lemma. Now, remark that the sign changes of G in \((0,2m+1)\) occur at integer points \(2,3,\ldots 2m\) and in \(\frac{1}{2}\) (by symmetry). Thus, we have

$$\begin{aligned} J := \displaystyle {\int _1^{M}} |G(x)|\,\mathrm{d}x= & {} \displaystyle {\sum _{k=1}^{2m} (-1)^k \int _k^{k+1}} G(x)\,\mathrm{d}x \\= & {} \displaystyle {2 \sum _{k=0}^{2m-1} (-1)^k \int _k^{k+1}F(x)\,\mathrm{d}x }, \end{aligned}$$

since for all \(x \in [0,M]\), we have \(G(x) = F(x)-F(x-1)\) and (because M is odd) \(F(x) = F(M-x)\). It follows that

$$\begin{aligned} J \ge \displaystyle {2 \left( \sum _{k=0}^{2m} (-1)^k \int _k^{k+1}F(x)\,\mathrm{d}x \right) }-2, \end{aligned}$$

since \(|F| \le 1\) everywhere.

Consequently, by isolating the index \(\alpha =0\) in the definition of F, we get \(J \ge 2\left( J^\prime +\frac{1}{M}\right) -2\), with

$$\begin{aligned} J^\prime = \sum _{k=0}^{2m} \frac{ (-1)^k }{M} \sum \limits _{\begin{array}{c} -m\le \alpha \le m \\ \alpha \ne 0 \end{array}} \int _k^{k+1} \mathrm{e}^{2i\omega \alpha x} \,\mathrm{d}x. \end{aligned}$$

By exchanging the sums and grouping identical terms, we obtain

$$\begin{aligned} J^\prime= & {} \displaystyle {\frac{1}{M} \sum \limits _{\begin{array}{c} -m\le \alpha \le m \\ \alpha \ne 0 \end{array}} \sum _{k=0}^{2m} (-1)^k \cdot \frac{\mathrm{e}^{2i\omega \alpha (k+1)}- \mathrm{e}^{2i\omega \alpha k} }{2i\omega \alpha }} \nonumber \\= & {} \displaystyle {\sum \limits _{\begin{array}{c} -m\le \alpha \le m \\ \alpha \ne 0 \end{array}} \frac{-1}{i\pi \alpha } \sum _{k=1}^{2m} \left( -\mathrm{e}^{2i\omega \alpha }\right) ^k}. \end{aligned}$$
(76)

After summation of the geometric progression

$$\begin{aligned}&\displaystyle {\sum _{k=1}^{2m}} \left( -\mathrm{e}^{2i\omega \alpha }\right) ^k = -\mathrm{e}^{2i\omega \alpha }\cdot \dfrac{1-\mathrm{e}^{2i\omega \alpha (2m)}}{1+\mathrm{e}^{2i\omega \alpha }} \\&\quad = \mathrm{e}^{i\pi \alpha } \dfrac{i \sin (2\omega m\alpha )}{\cos (\omega \alpha )} = \dfrac{i \sin (2\omega m\alpha -\pi \alpha )}{\cos (\omega \alpha )} = -i\tan (\omega \alpha ), \end{aligned}$$

Equation (76) finally leads to

$$\begin{aligned} J^\prime = \sum \limits _{\begin{array}{c} -m\le \alpha \le m \\ \alpha \ne 0 \end{array}} \frac{1}{\pi \alpha } \cdot \tan (\omega \alpha ) = \frac{2}{M} \sum _{\alpha =1}^m g(\omega \alpha ) \end{aligned}$$

where \(g= t \mapsto \frac{\tan t}{t}\). Now since g is positive and increasing on \((0,\frac{\pi }{2})\), we have

$$\begin{aligned} \sum _{\alpha =1}^m g(\omega \alpha ) \ge \int _0^m g(\omega x) \,\mathrm{d}x = \frac{1}{\omega } \int _0^{\omega m} g(t)\,\mathrm{d}t. \end{aligned}$$

Using the lower bound \(g(t)\ge \frac{2}{\pi } \tan t\) for \(t\in (0,\frac{\pi }{2})\), we finally get

$$\begin{aligned} J^\prime\ge & {} \frac{4}{\pi ^2} \int _0^{\omega m} \! \tan t\,dt = -\frac{4}{\pi ^2} \log \cos (\omega m)\\= & {} -\frac{4}{\pi ^2}\log \sin \left( \frac{\omega }{2} \right) \end{aligned}$$

and thus \(\displaystyle J^\prime \ge \frac{4}{\pi ^2} \log \left( \frac{2}{\omega } \right) \), from which the inequality announced in Lemma 1 follows. \(\square \)

Now, let us prove the Theorem 3 by building a discrete image u such that \({\textsc {STV}}_1(u)\) is fixed but \({\textsc {STV}}_\infty (u)\) increases with the image size. We consider the function H defined by

$$\begin{aligned} \forall x \in {\mathbb R}, \quad H(x) = \int _0^x G(t)\,\mathrm{d}t, \end{aligned}$$

where \(G \in T_M\) is the real-valued M-periodic trigonometric polynomial defined in Lemma 1 (\(M=2m+1\)). Since the integral of G over one period is zero (\(\int _0^M G(t) \, dt = 0\)), H is also an element of \(T_M\). Consequently, the bivariate trigonometric polynomial defined by

$$\begin{aligned} \forall (x,y) \in {\mathbb R}^2, \quad U(x,y) = \frac{1}{M} H(x), \end{aligned}$$

belongs to \(T_M \otimes T_M\), and since M is odd it is exactly the Shannon interpolate of the discrete image defined by

$$\begin{aligned} \forall (k,l) \in I_M\times I_M, \quad u(k,l) = U(k,l). \end{aligned}$$
(77)

In particular, by definition of \({\textsc {STV}}_1\) and \({\textsc {STV}}_\infty \), we have

$$\begin{aligned} {\textsc {STV}}_1(u)= & {} \sum _{(k,l) \in \Omega } |\nabla U(k,l)|, \quad \\ \text {and} \quad {\textsc {STV}}_\infty (u)= & {} \int _{[0,M]^2} |\nabla U(x,y)| \, \mathrm{d}x \mathrm{d}y. \end{aligned}$$

From Lemma 1, we have on the one hand,

$$\begin{aligned} {\textsc {STV}}_1(u)= & {} \sum _{(k,l) \in \Omega } |\nabla U(k,l)| \\= & {} \sum _{k=0}^{2m} |H'(k)| = \sum _{k=0}^{2m} |G(k)|= 2, \end{aligned}$$

and on the other hand,

$$\begin{aligned} {\textsc {STV}}_\infty (u)= & {} \int _{[0,M]^2} |\nabla U(x,y)| \, \mathrm{d}x\mathrm{d}y = \int _0^M |H'(x)|\,\mathrm{d}x \\= & {} \int _0^M |G(x)|\,\mathrm{d}x \ge \frac{8}{\pi ^2} \log \left( \frac{2M}{\pi }\right) -2. \end{aligned}$$

which cannot be bounded from above by a constant independent of M. \(\square \)

Appendix 5: Proof of Proposition 10

Let \(u \in {\mathbb R}^{\Omega }, n \in {\mathbb N}\) and \(\alpha \in {\mathbb R}\) such that \(n \ge 1\) and \(\alpha > 0\). One can rewrite \({\textsc {HSTV}}_{\alpha ,n}(u) = \tfrac{1}{n^2} H_\alpha (\nabla _nu)\), where

$$\begin{aligned} \forall g \in {\mathbb R}^{\Omega _n} \times {\mathbb R}^{\Omega _n}, \quad H_\alpha (g) = \sum _{(x,y) \in \Omega _n} \mathcal {H}_{\alpha }(g(x,y)). \end{aligned}$$

Let us show that the Legendre–Fenchel transform of \(H_\alpha \) is

$$\begin{aligned} H_\alpha ^\star (p)= \delta _{\Vert \cdot \Vert _{\infty ,2} \le 1}(p) + \tfrac{\alpha }{2} \Vert p\Vert _2^2. \end{aligned}$$

One easily checks that \(\mathcal {H}_\alpha \in \Gamma ({\mathbb R}^2)\), and it follows that \(H_\alpha \in \Gamma ({\mathbb R}^{\Omega _n} \times {\mathbb R}^{\Omega _n})\). Thus, for any image \(u\in {\mathbb R}^\Omega \), we have \(H_\alpha (\nabla _nu) = H_\alpha ^{\star \star }(\nabla _nu)\) and

$$\begin{aligned} H_\alpha ^{\star \star }(\nabla _nu) = \sup _{p \in {\mathbb R}^{\Omega _n} \times {\mathbb R}^{\Omega _n}} \langle \nabla _nu, p \rangle - H_\alpha ^\star (p). \end{aligned}$$
(78)

Besides, we have \(H_\alpha ^\star (p) = \sum _{(x,y) \in \Omega _n} \mathcal {H}^\star _\alpha (p(x,y))\), and the Legendre–Fenchel transform of \(\mathcal {H}_\alpha \) is the function \(\mathcal {H}_\alpha ^\star (z) = \delta _{|\cdot | \le 1}(z) + \tfrac{\alpha }{2}|z|^2\), where \(\delta _{|\cdot | \le 1}\) denotes the indicator function of the unit ball for the \(\ell ^2\) norm in \({\mathbb R}^2\). Indeed, it is proven in [55] that \(\mathcal {H}_\alpha \) is the Moreau envelope (or Moreau–Yosida regularization) [51, 76] with parameter \(\alpha \) of the \(\ell ^2\) norm \(|\cdot |\), or equivalently the infimal convolution (see [59]) between the two proper, convex and l.s.c functions \(f_1(x) = |x|\) and \(f_2(x) = \tfrac{1}{2\alpha } |x|^2\), that is

$$\begin{aligned} \forall y \in {\mathbb R}^2, \quad \mathcal {H}_\alpha (y) = \left( f_1 \Box f_2 \right) (y) := \inf _{x \in {\mathbb R}^2} f_1(x) + f_2(y-x). \end{aligned}$$

Thus, we have \(\mathcal {H}_\alpha ^\star = \left( f_1 \Box f_2 \right) ^\star = f_1^\star + f_2^\star \) (see [55, 59]), leading exactly to \(\mathcal {H}_\alpha ^\star (z) = \delta _{|\cdot | \le 1}(z) + \tfrac{\alpha }{2}|z|^2 \) for any \(z \in {\mathbb R}^2\), since we have \(f_1^\star = z \mapsto \delta _{|\cdot |\le 1}(z)\) and \(f_2^\star = z \mapsto \tfrac{\alpha }{2} |z|^2\). It follows that for any \(p \in {\mathbb R}^{\Omega _n} \times {\mathbb R}^{\Omega _n}\), we have

$$\begin{aligned} H_\alpha ^\star (p) = \! \sum _{(x,y) \in \Omega _n} \!\! \mathcal {H}^\star _\alpha (p(x,y)) = \delta _{\Vert \cdot \Vert _{\infty ,2} \le 1}(p) + \tfrac{\alpha }{2}\Vert p\Vert _2^2,\nonumber \\ \end{aligned}$$
(79)

and the supremum (78) is a maximum for the same reason as in the proof of Proposition 9. Finally, writing \({\textsc {HSTV}}_{\alpha ,n}(u) = \tfrac{1}{n^2} H_\alpha (\nabla _nu) = \tfrac{1}{n^2} H_\alpha ^{\star \star }(\nabla _nu)\) using (78) and (79) leads to the announced result. \(\square \)

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Abergel, R., Moisan, L. The Shannon Total Variation. J Math Imaging Vis 59, 341–370 (2017). https://doi.org/10.1007/s10851-017-0733-5

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Keywords

  • Total variation
  • Image interpolation
  • Shannon theory
  • Legendre–Fenchel duality
  • Aliasing
  • Image restoration