Unique bounded operator constructed from sequence of given operators from orthogonal sum of Hilbert spaces to another Hilbert space.

Question

Let $(H_k)_{k\in\mathbb{N}}$ be a sequence of Hilbert spaces. Define

$$ \oplus_{k\in\mathbb{N}}H_k:=\{(x_k)_k \in\Pi_{k\in\mathbb{N}}H_k \mid \sum_{k\in\mathbb{N}}||x_k||^2<\infty\} $$ with $$ \langle(x_k)_k, (y_k)_k\rangle := \sum_{k\in\mathbb{N}}\langle x_k,y_k\rangle. $$

This is direct orthogonal sum is again a Hilbert Space.

I then need to show that for every sequence of operators $T_k\in L(H_k,F)$ (linear and bounded operators) with $F$ another Hilbert Space with the property

$$ \sum_{k\in\mathbb{N}} ||T_kx_k||^2 < \infty \text{ for all } (x_k)_k \in \oplus_{k\in\mathbb{N}} H_k $$

that we have that there exists a unique bounded operator

$$ T: \oplus_{k\in\mathbb{N}}H_k\to F $$

with the property that $T_{\mid H_k}=T_k$ for all $k\in\mathbb{N}$.

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Now I didn't really know where to start, since the chapter is on self adjoint operators, and I don't see them here (all the operators do not map back to the same space, so they cannot be self adjoint).

I started to prove uniqueness:

Suppose there exist $T$ and $T'$ with this property. Then $T_{|H_k}=T_k=T'_{|H_k}$ for all $k\in\mathbb{N}$. But this seems not done yet.

Now I don't know how to use the property we know about the $T_k$. It looks similar to the $l^2$ space though, and I think it might be necessary to know that $T$ will be bounded.

Since all the $T_k$ map to $F$, but also $T$ maps to $F$, we must have that $T$ is the sum or product of the $T_k$ applied to an $(x_k)_k$:

Something like: $T((x_k)_k) := \sum T_{|H_k}x_k = \sum T_kx_k$. Then this would indeed be an element of $F$ right? If it would be defined like this, I think my uniqueness argument would be complete, since $T$ only consists of the $T_k$ which will be the same for $T$ and $T'$. But how do I know it must be defined like this? And not instead, for example, using a product symbol instead of a sum symbol?

Then for boundedness I think I need to start with looking at $||\sum_{k\in\mathbb{N}}T_kx_k||^2$, but I do not know how to continue.

Answer 1

As Ionza remarked, it’s important to make precise what we mean by $T|_{H_k} = T_k$. For $x = (x_k) \in \oplus_k H_k$, let $x^\ell$ be the sequence $$x_k^\ell = \begin{cases} x_\ell & k = \ell \\ 0 & k \neq \ell \end{cases}$$ That is $x^\ell$ is the sequence with $x_\ell$ in the $\ell$ position and $0$ everywhere else.

When we say $T|_{H_{k_0}} = T_{k_0}$, we mean that if $x \in \oplus_k H_k$ such that $x_k = 0$ for all $k \neq k_0$, then $T x = T_{k_0} x_{k_0}$.

Now, suppose $T’ : \oplus_k H_k \to F$ is a bounded linear operator such that $T’|_{H_k} = T_k$ for all $k$. Our first observation is that $T’ x^\ell = T_\ell x_\ell$.

Next, we deduce that $T’$ is uniquely determined for $x$ such that $x_k = 0$ for all $k > N$, for some $N$. We can first observe that $x = \sum_{k = 1}^N x^k$, so if $T’$ is linear, then \begin{align*} T’x & = T’ \sum_{k = 1}^{N} x^k \\ & = \sum_{k = 1}^N T’ x^k \\ & = \sum_{k = 1}^N T_k x_k \end{align*} Thus we’ve shown that $T’$ is uniquely determined for $x$ such that $x_k = 0$ for all but finitely many $k$.

Our last step to establish uniqueness is to extend to all of $\oplus_k H_k$. To do this, we need to establish that the set of $x$ such that $x_k = 0$ for all but finitely many $k$ is dense in $\oplus_k H_k$. For any $x \in \oplus_k H_k$, consider the sequence $\left( \sum_{k = 1}^N x^k \right)_{N = 1}^{\infty}$. Then \begin{align*} \left\| x - \sum_{k = 1}^N x^k \right\| & = \sqrt{\sum_{ k = N + 1}^{\infty} \| x_k \|^2 } \\ & \stackrel{N \to \infty}{\to} 0 , \end{align*} since $\oplus_k H_k$ is the set of all sequences where $\sum_{k = 1}^{\infty} \|x_k\|^2 < \infty$.

To turn this density fact into our final uniqueness statement, we note that $T’$ is bounded iff it’s continuous, so \begin{align*} T’ x & = T’ \lim_{N \to \infty} \sum_{k = 1}^N x^k \\ & = \lim_{N \to \infty} T’ \sum_{k = 1}^N x^k \\ & = \lim_{N \to \infty} \sum_{k = 1}^N T_k x_k \\ & = \sum_{k = 1}^{\infty} T_k x_k . \end{align*}

This concludes the proof of uniqueness.

The proof of existence is relatively straightforward. The condition that $\sum_k \| T_k x_k \|^2 < \infty$ ensures that the operator is well-defined, i.e. that the series $\sum_{k = 1}^\infty T_k x_k$ actually converges (check it). It remains to show that it’s bounded. This will follow from the uniform boundedness principle. If we define $$S_N x = \sum_{k = 1}^N T_k x_k $$ then $T$ is the pointwise limit of the bounded operators $S_N$.

EDIT: To see why the condition that $\sum_k \| T_k x_k \|^2 < \infty$ is needed, consider the following example. Let $H_k = \mathbb{R}$ for all $k$ and $F = \mathbb{R}$. Let $$T_k t = k t .$$ Consider $x = (1/k)_k$. Then $\sum_{k = 1}^{\infty} \| x_k \|^2 = \sum_{k=1}^\infty k^{-2} < \infty$, but $\sum_{k = 1}^\infty \| T_k x_k \|^2 = \sum_{k = 1}^{\infty} 1 = \infty$. If we try to evaluate $T x$, we get $$Tx = \sum_{k = 1}^\infty T_k x_k = \sum_{k = 1}^\infty k (1/k) = \infty .$$ Thus $T$ is not well-defined, because it gives us nonsense.

Answer 2

I think the problem is that $\ H_k\ $, strictly speaking, isn't actually a subspace of $\ \bigoplus_{k\in\mathbb{N}}H_k\ $, so to make sense of $\ T_{|H_k}\ $ you need to choose an appropriate subspace $\ \mathcal{H}_k\ $ of $\ \bigoplus_{k\in\mathbb{N}}H_k\ $ which you can take as being, in some sense, the "same" as $\ H_k\ $, and to make sense of $\ T_{|\mathcal{H}_k} =T_k\ $, you have to interpret it as meaning $\ T_{|\mathcal{H}_k}(x)=T_k x_k\ $ for all $\ x\in \mathcal{H}_k\ $. The obvious candidate for $\ \mathcal{H}_k\ $ would appear to be $$ \mathcal{H}_k\stackrel{\text{Def}}{=}\left\{x\in\bigoplus_{j\in\mathbb{N}}H_j\,\bigg|\,x_j=0\ \text{ for }\ j\ne k\right\}\ $$ —that is, the projection of $\ \bigoplus_{k\in\mathbb{N}}H_k\ $ onto its $\ k^\text{th}\ $ coordinate. The identification of $\ H_k\ $ with $\ \mathcal{H}_k\ $ is somewhat analogous to the identification of the real numbers with the subset of the complex numbers whose imaginary parts are zero.

Now I think you should be able to show that the only operator $\ T: \bigoplus_{j\in\mathbb{N}}H_j\rightarrow F\ $ for which $\ T_{|\mathcal{H}_k}(x)=T_k x_k\ $ is given by your definition $\ T(x)=\sum_{k\in\mathbb{N}}T_kx_k\ $.