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Let $T:H_1 \rightarrow H_2$, where $H_1$ and $H_2$ are real hilbert spaces and $T$ is a bounded linear operator. Prove the following:

suppose $\{e_j\}$ an orthonormal basis for $H_2$, show that there exist a sequence $\{f_j\}$ in $H_1$ such that for all $x \in H_1$:

$T(x) = \sum \langle x,f_j\rangle_1 e_j$

My guess:

I know that because $\{e_j\}$ is an orthonormal basis, we get $T(x) = \sum \langle T(x),e_j\rangle_2 e_j$. So we need to find a sequence $\{f_j\}$ such that:

$\langle x,f_j\rangle_1 = \langle T(x),e_j\rangle_2$.

I think i need to use Riesz-Frechet but i dont see how...

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  • $\begingroup$ if $H_2 = H_1^{'}$ this follows from Riesz Frechet but we cant assume $H_2 = H_1^{'}$ right? $\endgroup$
    – Kees Til
    Jul 19 '16 at 7:35
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For each $j$, the map $x\mapsto \langle Tx,e_j\rangle_2$ is a bounded linear map from $H_1$ to $\mathbb R$, so by Fréchet–Riesz there exists $f_j\in H_1$ such that $\langle Tx,e_j\rangle_2 = \langle x,f_j\rangle_1$.

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  • $\begingroup$ nice answer, how do you see that the map is a B.L.O? Is it because it's obviously a constant function? Or can u use the definition as well: $||T(x)|| \leq k \cdot ||x||$ $\endgroup$
    – Kees Til
    Jul 19 '16 at 7:48
  • $\begingroup$ @Kees: It is not constant unless the range of $T$ is orthogonal to $e_j$, in which case it would be $0$. It is linear because $T$ is linear and the inner product is linear in each variable, and it is bounded because $|\langle Tx,e_j\rangle|\leq \|Tx\|\|e_j\|=\|Tx\|\leq \|T\|\|x\|$. $\endgroup$ Jul 19 '16 at 7:52
  • $\begingroup$ But if you choose $j$ as a constant number, say $2$, the inproduct is constant right? Like for example $<T(x),e_2> = k$. You choose each $f_j$ depending on the $j$, so each of these functions should be constant right or can you think of a counterexample? $\endgroup$
    – Kees Til
    Jul 19 '16 at 8:03
  • $\begingroup$ nevermind, this is silly because $T(x)$ is still a variable that depends on $x$. Thanks for your help :D $\endgroup$
    – Kees Til
    Jul 19 '16 at 8:04
  • $\begingroup$ @KeesTil: Yes, $f_j$ depends only on $j$, but the map is a function of $x$, and varies with $x$, unless it is the $0$ map (in which case $f_j$ would be the $0$ element). $\endgroup$ Jul 19 '16 at 8:17
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Define the sequence $T_j:H_1\to \mathbb{R}$ by $T_j(x) = \langle T(x),e_j\rangle_2$.

Note that each element in the sequence is a bounded linear function because:

a. $T_j(x+y) = \langle T(x+y),e_j\rangle_2=\langle T(x)+T(y), e_j\rangle_2 = \langle T(x),e_j\rangle_2 + \langle T(y), e_j\rangle_2$ $= T_j(x) + T_j(y)$.

This is because $T$ is linear and $\langle \cdot, \cdot \rangle_2$ is an inner product.

b. For any scalar $c$ we have $T_j(cx) = \langle T(cx),e_j\rangle_2 = \langle cT(x), e_j\rangle_2 = c\langle T(x),e_j\rangle_2 = cT_j(x)$.

This is due to the same reasons as above.

c. Since $T$ is a bounded linear operator we have that there exists a $K\in \mathbb{R}$ such that for every $x\in H_1$ we have

$$\|T(x)\|_2 \leq K\|x\|_1.$$

Using the operator norm this gives

$$\|T_j\|_{op} \leq \sup_{\|x\|_1\leq 1}\|T(x)\|_2 \leq \sup_{\|x\|_1\leq 1} K\|x\|_1 \leq K.$$

These three requirements shows that $T_j\in H_1^*$ (the dual space) for every $j$. Thus, by the Riesz representation theorem there exists an $f_j\in H_1$ such that

$$T_j(x) := \langle T(x),e_j\rangle_2 = \langle x, f_j\rangle_1$$

for every $x\in H_1$ and fixed $j$. We conclude that

$$T(x) = \sum_j \langle T(x),e_j\rangle_2 e_j = \sum_j T_j(x) e_j = \sum_j \langle x, f_j \rangle_1 e_j$$

as desired.

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  • $\begingroup$ It's a slly typo, you are correct. It has been edited. $\endgroup$
    – William
    Jul 19 '16 at 8:33
  • $\begingroup$ Ah, I see. Yes, another silly typo. It should be an inequality (not equality) from Cauchy - Shwarz: $|\langle T(x), e_j \rangle_2| \leq \|T(x)\|_2$ $\endgroup$
    – William
    Jul 19 '16 at 8:39

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