Let us give a quick summary of what it means for a function to be holomorphic on a Banach space; we refer the reader to the book [6] by Dineen for a comprehensive treatment. Let D be an open subset of a Banach space X, and \(f:D \rightarrow Y\) a map into a Banach space Y. We say f has a Gâteaux derivative at x if
$$\begin{aligned} \lim _{\lambda \rightarrow 0} \frac{f(x + \lambda h) - f(x)}{\lambda } \overset{\mathrm{def}}{=} Df(x)[h] \end{aligned}$$
exists for all \(h \in X\). If f has a Gâteaux derivative at every point of D it is Gâteaux holomorphic [6, Lemma 3.3], i.e. holomorphic on each one dimensional slice. If, in addition, f is locally bounded on D, then it is actually Fréchet holomorphic [6, Proposition 3.7], which means that for each x there is a neighborhood G of 0 such that the Taylor series
$$\begin{aligned} f(x+h) = f(x) + \sum _{k=1}^\infty D^k f(x) [h,\ldots , h], \qquad h \in G, \end{aligned}$$
(7)
converges uniformly for all h in G. The kth derivative is a continuous linear map from \(X^k \rightarrow Y\), which is evaluated on the k-tuple \((h,h, \ldots , h)\).
The following lemma is the IP version of [8, Proposition 2.5] and [13, Proposition 2.2].
Lemma 3.1
Let D be an open set in \({B(\mathscr {H})}^d\) that is closed with respect to countable direct sums, and let \(F :D \rightarrow {B(\mathscr {H})}\) be intertwining preserving. Then F is bounded on bounded subsets of D, continuous and Gâteaux differentiable.
Proof
(Locally bounded) Suppose there were \(x_n \in D\) such that \(\{ \Vert x_n \Vert \} \) is bounded, but \(\{ \Vert F(x_n) \Vert \}\) is unbounded. Since D is closed with respect to countable direct sums, there exists some unitary \(u :\mathscr {H}\rightarrow \mathscr {H}^\infty \) such that \(u^* ( \bigoplus x_n ) u \in D\). Since F is IP, by Definition 1.2, we have \( [\bigoplus F(x_n)]\) is bounded, which is a contradiction.
(Continuity) Fix \(a \in D \) and let \(\varepsilon > 0\). By hypothesis, there exists a unitary \(u :\mathscr {H}\rightarrow \mathscr {H}^2\) such that
$$\begin{aligned} \alpha = \ u^* \begin{bmatrix} a&\quad 0 \\ 0&\quad a \end{bmatrix} u \in D. \end{aligned}$$
(8)
Choose \(\delta _1 > 0\) such that \(B( a , \delta _1) \subseteq D\), \(B( \alpha , \delta _1) \subseteq D\), and such that on \(B( \alpha , \delta _1)\) the function F is bounded by M. Choose \(\delta _2 > 0\) such that \(\delta _2 < \min ( \delta _1/2, \varepsilon \delta _1/2M)\). Note that for any \(a,b \in {B(\mathscr {H})}^d\) and any \(\lambda \in \mathbb {C}\), we have
$$\begin{aligned} u^* \begin{bmatrix} I&\quad - \lambda \\ 0&\quad I \end{bmatrix} \begin{bmatrix} b&\quad 0 \\ 0&\quad a \end{bmatrix} \begin{bmatrix} I&\quad \lambda \\ 0&\quad I \end{bmatrix} u = u^* \begin{bmatrix} b&\quad \lambda (b-a) \\ 0&\quad a \end{bmatrix} u. \end{aligned}$$
(9)
So by part (ii) of the definition of IP (Definition 1.2) we get that if \(\Vert b - a \Vert < \delta _2\), and letting \(\lambda = M/\varepsilon \), then
$$\begin{aligned} F\left( u^* \begin{bmatrix} I&\quad - M/\varepsilon \\ 0&\quad I \end{bmatrix} \begin{bmatrix} b&\quad 0 \\ 0&\quad a \end{bmatrix} \begin{bmatrix} I&\quad M/\varepsilon \\ 0&\quad I \end{bmatrix} u \right) = u^* \begin{bmatrix}F( b)&\;\;M[F (b)-F(a)]/\varepsilon \\ 0&\;\;F(a) \end{bmatrix} u \end{aligned}$$
is bounded by M. In particular, since the norm of the (1, 2)-entry of the last matrix is bounded by the norm of the whole matrix, we see that \(\Vert M(F(b)-F(a))/\varepsilon \Vert < M\), so \(\Vert F(b)-F(a) \Vert < \varepsilon \).
(Differentiability) Let \(a \in D\) and \(h \in {B(\mathscr {H})}^d\). Let u be as in (8). Choose \(\varepsilon > 0\) such that, for all complex numbers t with \(|t| < \varepsilon \),
$$\begin{aligned} u^* \begin{bmatrix} a + th&\;\;\varepsilon h \\ 0&\;\;a \end{bmatrix} u \in D, \end{aligned}$$
and \(a + th \in D\). Let \(b = a +th\) and \(\lambda =\varepsilon /t\) in (9), and as before we conclude that
$$\begin{aligned} F\left( u^* \begin{bmatrix} a + th&\;\; \varepsilon h \\ 0&\;\;a \end{bmatrix} u \right) = u^* \begin{bmatrix} F(a + th)&\;\;\varepsilon (F(a+th) - F(a))/t \\ 0&\;\;F(a) \end{bmatrix} u . \end{aligned}$$
(10)
As F is continuous, when we take the limit as \(t \rightarrow 0\) in (10), we get
$$\begin{aligned} F\left( u^* \begin{bmatrix} a&\quad \varepsilon h \\ 0&\quad a \end{bmatrix} u \right) = u^* \begin{bmatrix} F(a )&\quad \varepsilon DF (a)[h] \\ 0&\quad F(a) \end{bmatrix} u . \end{aligned}$$
Therefore \(DF(a) [h]\) exists, so F is Gâteaux differentiable, as required.\(\square \)
When we replace X by a Banach algebra (in our present case, this is \({B(\mathscr {H})}^d\) with coordinate-wise multiplication), we would like something more than Fréchet holomorphic: we would like the kth term in (7) to be an actual free polynomial, homogeneous of degree k, in the entries of h.
The following result was proved by Kaliuzhnyi-Verbovetskyi and Vinnikov [13, Theorem 6.1] and by Klep and Špenko [14, Proposition 3.1].
Theorem 3.2
Let
$$\begin{aligned} g:{\mathbb M}^{[d]}\rightarrow \mathscr {M}^1, \qquad x\mapsto g(x) \end{aligned}$$
be an nc-function such that each matrix entry of g(x) is a polynomial of degree less than or equal to N in the entries of the matrices \(x^r\), \(1 \le r \le d\). Then g is a free polynomial of degree less than or equal to N.
We extend this result to multilinear SSOC IP maps. Each \(h_j\) will be a d-tuple of operators, \((h_j^1, \ldots , h_j^d)\).
Proposition 3.3
Let
$$\begin{aligned} L:{B(\mathscr {H})}^{dN} \rightarrow {B(\mathscr {H})}, \qquad (h_1, \ldots , h_N) \mapsto L(h_1,\ldots , h_N) \end{aligned}$$
be a continuous N-linear map from \(({B(\mathscr {H})}^{d})^N\) to \({B(\mathscr {H})}\) that is IP and SSOC. Then L is a homogeneous polynomial of degree N in the variables \(h_1^1, \ldots , h_N^d\).
Proof
By Proposition 2.3, if we restrict L to \({\mathscr {M}}^{dN}\), we get an nc-function. By Theorem 3.2, there is a free polynomial p of degree N that agrees with L on \({\mathscr {M}}^{dN}\). By homogeneity, p must be homogeneous of degree N. Define
$$\begin{aligned} {\Delta }(h) = L(h) - p(h) . \end{aligned}$$
Then \({\Delta }\) vanishes on \(({\mathscr {M}}^{d})^{N}\), and is SSOC. Since \(({\mathscr {M}}^{d})^{N}\) is strong operator topology dense in \(({B(\mathscr {H})}^{d})^N\), it follows that \({\Delta }\) is identically 0. \(\square \)
One of the achievements of Kaliuzhnyi-Verbovetskyi and Vinnikov in [13] is the Taylor–Taylor formula [13, Theorem 4.1]. This comes with a remainder term, which can be estimated. They show [13, Theorem 7.4] that with the assumption of local boundedness, this renders an nc-function analytic. The following theorem is an IP version of the latter result.
Theorem 3.4
Let D be an open neighborhood of 0 in \({B(\mathscr {H})}^d\), and let \(F :D \rightarrow {B(\mathscr {H})}\) be a function that is intertwining preserving and sequentially strong operator continuous. Then there is an open set \(U \subseteq D\) containing 0 and homogeneous free polynomials \(P_k\) of degree k such that
$$\begin{aligned} F(x) = F(0) + \sum _{k=1}^\infty P_k(x), \qquad x \in U, \end{aligned}$$
where the convergence is uniform for \(x \in U\).
Proof
Any open ball centered at 0 is closed with respect to countable direct sums, so we can assume without loss of generality that D is closed with respect to countable direct sums and bounded. By Lemma 3.1, F is bounded and Gâteaux differentiable on D, and so by [6, Proposition 3.7], F is automatically Fréchet holomorphic. Therefore, there is some open ball U centered at 0 such that
$$\begin{aligned} F(h) = F(0) + \sum _{k=1}^\infty D^k F(0)[h,\ldots , h], \qquad h \in U . \end{aligned}$$
We must show that each \(D^k F(0) [h,\ldots , h]\) is actually a free polynomial in h.
Claim 3.5
For each \(k \in \mathscr {N}\), the function
$$\begin{aligned} G^k:\big (h^0, \ldots , h^k\big ) \mapsto D^k F(h^0) \big [h^1,\ldots , h^k\big ] \end{aligned}$$
(11)
is an IP function on \( U {\times } ({B(\mathscr {H})}^d)^k \subseteq ({B(\mathscr {H})}^d)^{k+1}\).
Proof
Indeed, when \(k=1\), we have
$$\begin{aligned} D F\big (h^0\big ) \big [h^1\big ] = \lim _{t \rightarrow 0} \frac{1}{t} \,\Bigl [ F\big (h^0{ +} th^1\big ) - F\big (h^0\big )\Bigr ]. \end{aligned}$$
(12)
As F is IP, so is the right-hand side of (12). For \(k > 1\),
$$\begin{aligned} D^k F(h^0) [h^1\!,\dots , h^k] = \lim _{t \rightarrow 0}&\frac{1}{t} \,\bigl [D^{k-1} F(h^0{+}th^k) [h^1\!, \dots , h^{k-1}]\\&\qquad \qquad - D^{k-1}F(h^0) [h^1\!,\dots , h^{k-1}] \bigr ]. \end{aligned}$$
By induction, these are all IP. \(\blacksquare \)
Claim 3.6
For each \(k \in \mathscr {N}\), the function \(G^k\) from (11) is SSOC on \( U {\times } ({B(\mathscr {H})}^d)^k\).
Proof
Again we do this by induction on k. Let \(G^0 = F\), which is SSOC on \(U \subseteq D\) by hypothesis. Since \(G^{k-1}\) is IP on the set \(U^k\), it is locally bounded, and by Lemma 3.1 it is Gâteaux differentiable. Suppose
$$\begin{aligned} \mathrm{SOT} \lim _{n \rightarrow \infty } h^j_n = h^j, \qquad 0 \le j \le k , \end{aligned}$$
where each \(h^j_n\) and \( h^j \) is in U. Let h denote the \((k{+}1)\)-tuple \( (h^0, \ldots , h^{k})\) in \(U^{k+1}\), and let \({\widetilde{h}}\) denote the k-tuple \( (h^0, \ldots , h^{k-1})\); similarly, let \(h_n\) denote \( (h_n^0, \ldots , h_n^{k})\) and \({\widetilde{h}}_n\) denote \( (h_n^0, \ldots , h_n^{k-1})\). There exists some unitary u so that \( y = u^* \bigl (\widetilde{h} {\oplus } {\widetilde{h}}_1 {\oplus } {\widetilde{h}}_2 {\oplus } \cdots \bigr ) u\) is in \(U^{k}\). Since \(G^{k-1}\) is differentiable at y, and is IP, we have that the diagonal operator with entries
$$\begin{aligned}&\displaystyle \frac{1}{t}\, \bigl [ G^{k-1}(h^0 {+} t h^k\!, h^1\, \dots , h^{k-1}) - G^{k-1}(h^0\! , h^1\!, \dots , h^{k-1})\bigr ], \nonumber \\&\displaystyle \frac{1}{t}\,\bigl [G^{k-1}(h^0_1 {+} t h^k_1, h^1_1, \dots , h^{k-1}_1) - G^{k-1}(h^0_1 , h^1_1, \dots , h^{k-1}_1) \bigr ], \\&\displaystyle \cdots \nonumber \end{aligned}$$
(13)
has a limit as \(t \rightarrow 0\).
Let \(\varepsilon > 0\), and let \(v \in \mathscr {H}\) have \(\Vert v \Vert \le 1\). Choose t sufficiently close to 0 that each of the difference quotients in (13) is within \(\varepsilon /3\) of its limit (which is \(G^k\) evaluated at the appropriate h or \(h_n\)). Let n be large enough so that
$$\begin{aligned}&\bigl \Vert \bigl [ G^{k-1}(h^0 {+} t h^k\!, h^1\!, \dots , h^{k-1}) - G^{k-1}(h^0_n {+} t h^k_n, h^1_n, \dots , h^{k-1}_n) \bigr ] v \bigr \Vert \\&\qquad \quad + \bigl \Vert \bigl [ G^{k-1}(h^0\! , h^1\!, \dots , h^{k-1}) - G^{k-1}(h^0_n , h^1_n, \dots , h^{k-1}_n) \bigr ] v \bigr \Vert \le \frac{\varepsilon t}{3} . \end{aligned}$$
Then
$$\begin{aligned} \Bigl \Vert \Bigl [ G^k \big (h^0, \ldots , h^k\big ) - G^k \big (h^0_n, \ldots , h^k_n \big ) \Bigr ] v \Bigr \Vert \le \varepsilon . \end{aligned}$$
So each \(G^k\) is SSOC on \(U^{k+1}\). As \(G^k\) is linear in the last k variables, it is SSOC on \( U {\times } ({B(\mathscr {H})}^d)^k\) as claimed. \(\blacksquare \)
Therefore for each k, the map
$$\begin{aligned} \big (h^1, \ldots , h^k\big ) \mapsto D^k F(0) \big [h^1,\ldots , h^k\big ] \end{aligned}$$
is a linear IP function that is SSOC in a neighborhood of 0, so by Proposition 3.3 is a free polynomial. \(\square \)