Every bounded linear operator $T$ between real Hilbert spaces is $T(x) = \sum \langle x,f_j\rangle\, e_j$ 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...
 A: 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 $\langle Tx,e_j\rangle_2 = \langle x,f_j\rangle_1$.
A: 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.
