Corollary

— *Suppose that the hilbertian space E is the direct sum of the orthogonal subspaces M and N. The canonical isomorphism g from $S(M) \otimes S(N)$ onto $S(E)$ (A, III, § 6, No. 6) extends uniquely to a hilbertian space isomorphism h from $\hat{S}(M) \hat{\otimes}_2 \hat{S}(N)$ onto $\hat{S}(E)$.

Let $(e_i)_{i \in I}$ (resp. $(f_j)_{j \in J}$) be an orthonormal basis of the hilbertian space M (resp. N) and let $M_0$ (resp. $N_0$) be the vector subspace of E generated by the vectors $e_i$ (resp. $f_j$). Put $E_0 = M_0 + N_0$ and let $g_0$ be the canonical isomorphism from $S(M_0) \otimes S(N_0)$ onto $S(E_0)$. Put

for $\alpha \in \mathbf{N}^{(I)}$ and $\beta \in \mathbf{N}^{(J)}$. By prop. 4, we have thus defined the orthonormal bases $(z_\alpha)_{\alpha \in \mathbf{N}^{(I)}}$ for $\hat{S}(M)$, $(t_\beta)_{\beta \in \mathbf{N}^{(J)}}$ for $\hat{S}(N)$ and $(z_\alpha t_\beta)_{\alpha \in \mathbf{N}^{(I)}, \beta \in \mathbf{N}^{(J)}}$ for $\hat{S}(E)$. Since we have $z_\alpha t_\beta = g_0(z_\alpha \otimes t_\beta)$, and since the elements $z_\alpha \otimes t_\beta$ form an orthonormal basis of $\hat{S}(M) \hat{\otimes}_2 \hat{S}(N)$ (V, p. 29, cor. 1), we see that $g_0$ extends to a hilbertian space isomorphism $h : \hat{S}(M) \hat{\otimes}_2 \hat{S}(N) \to \hat{S}(E)$. By the construction, we have

for all vectors $x_1, ..., x_m$ in $M_0$ and $y_1, ..., y_n$ in $N_0$. By continuity, the same relation also holds for the vectors $x_1, ..., x_n$ in M and the vectors $y_1, ..., y_n$ in N; in other words, $h$ extends $g$. The uniqueness of $h$ is clear.

Let E and F be two hilbertian spaces and $u \in \mathscr{L}(E; F)$. The linear mapping $\hat{T}^n(u) = u \hat{\otimes}_2 ... \hat{\otimes}_2 u$ ($n$ factors) from $\hat{T}^n(E)$ into $\hat{T}^n(F)$ is continuous with norm $\|u\|^n$ (V, p. 28, formula (10)). Moreover, formulas (13) and (14) of V, p. 30, show that there exists an isomorphism $\phi_{n,E}$ from $\hat{S}^n(E)$ onto the subspace $\widehat{T S^n}(E)$ of $\hat{T}^n(E)$, and only one, such that

Hence there exists a continuous linear mapping $\hat{S}^n(u)$ from $\hat{S}^n(E)$ into $\hat{S}^n(F)$ and only one which makes the following diagram commutative

We now prove the formula

We clearly have $\| \hat{S}^n(u) \| \leq \| \hat{T}^n(u) \| = \| u \|^{n}$. Further, for all $x \in E$, we have $\hat{S}^n(u)(x^n) = u(x)^n, \| x^n \| = (n!)^{1/2} \| x \|^{n}$ and $\| u(x)^n \| = (n!)^{1/2} \| u(x) \|^{n}$, which gives

it follows immediately that $\| \hat{S}^n(u) \| \geq \| u \|^{n}$, hence formula (19).

It is clear that we have the formulas

Finally, $\hat{S}^n(u)$ coincides on $S^n(E)$ with the linear mapping $S^n(u) : S^n(E) \to S^n(F)$ defined in A, III, § 6, No. 2 since it transforms $x_1 ... x_n$ into $u(x_1) ... u(x_n)$ for every $x_1, ..., x_n$ in $E$.