Functional Analysis by Reed and Simon, chapter 3 exercise 15 Let $H$ be a Hilbert space with an orthonormal basis  $\{ x_n \}_{n=1}^\infty$ and let $\{ y_n \}_{n=1}^\infty$ be a sequence of elements in $H$. Show that following two statements are equivalent
$$ \forall x \in H , \ \left< x, y_n \right> \to 0, \ n \to \infty $$
$$ \forall m \in \mathbb{N}, \left< x_m, y_n \right> \to 0, \ n \to \infty \text{ and } \{ || y_n ||_H\}_{n=1}^\infty \text{ is bounded}$$
If we assume the first statement then clearly $x_m \in H$ so the first part of the second statement is shown. For the second part of the second statement define a family of operator of the form
$$ T_n (x) = \left< x, y_n \right>$$
Clearly the mapping is linear, and by Cauchy-Schwarz
$$ |T_n (x)| \leq || x ||_H || y_n||_H$$
So the operator is bounded and the norm of the operator is
$$ || T_n ||= || y_n ||_H $$
Applying the principle of uniform boundedness to this family of operators we find that
$$ \exists M > 0, \ \forall n \in \mathbb{N}, \  || T_n || \leq M$$
Since the operator norm coincided with the norm of each $y_n$ we may conclude that the sequence is bounded by $M$.
Assume the second statement. Fix $x \in H$ a computation shows
$$ \left< x, y_n \right> = \sum_{m=1}^\infty \left< x_m, x \right> \left< x_m, y_n \right>$$
Now I would like to "pass the limit" into the sum but I am unsure how to do this. We can also write $y_n$ in the form
$$y_n = \sum_{m=1}^\infty \left< x_m, y_n \right> x _m$$
Again "passing the limit" into the sum would actually give me that $y_n$ converges to $0$ but I cannot make either statement rigorous.
Edit : Here is an earlier idea I tried to implement. We have
$$| \left< x, y_n \right> | \leq \sum_{m=1}^\infty | \left< x_m, x \right> |  |\left< x_m, y_n \right> | $$
The series on the right converges, so its tail-end will converge to $0$. In particular for $\varepsilon > 0$ there exists $M > 0$ such that
$$ \sum_{m=1}^\infty | \left< x_m, x \right> |  |\left< x_m, y_n \right> | \leq \sum_{m=1}^{M - 1} | \left< x_m, x \right> |  |\left< x_m, y_n \right> | + \frac{\varepsilon}{2} $$
Now by the first hypothesis for $m \in \{1,...,M-1 \}$ there exists $n_m \in \mathbb{N}$ such that for $n \geq n_m$ we have
$$ |\left< x_m, y_n \right>| \leq \frac{1}{|\left< x_m, x \right>| + 1} \frac{\varepsilon}{2(M - 1)}$$
Now if we choose $N = \max \{ n_m : m \in \{ 1,..., M - 1 \}\}$ then for $n \geq N$ after a simple computation we have
$$ | \left< x, y_n \right> | \leq \varepsilon$$
I was hesitant to post this idea since I did not use the boundedness of the sequence $y_n$ anywhere.
 A: You have pointwise convergence of each term by the first condition. By the second condition condition there is a sequence converging to zero that majorizes the terms for arbitrary $ n $. This allows you to just split up the sum and show that both parts are close to zero.
EDIT: It is only a little bit trickier than what you now have and than what I sketched out above. In particular you need to be careful with your choice of the split point $M$.
First we are given $\epsilon>0$. Then we choose $M$ such that $$
\sum_{j=M+1}^\infty \langle x_m,x\rangle^2<\epsilon.
$$
(This is possible because the sequence of the squared coefficients is summable, indeed the sum equals the squared-norm of $x$)
Now we do what you did for the first $M$ terms and we bound the remainder via Cauchy-Schwarz:
$$
\left(\sum_{j=M+1}^\infty \langle x_m,x\rangle\langle x_m,y_n\rangle\right)^2 \leq \sum_{j=M+1} \langle x_m,x\rangle^2\sum_{m=M+1}\langle x_m,y_n\rangle^2\\ \leq \epsilon \|y_n\|^2
$$
(Note that the bound on the second factor comes again from the fact that the sum of squared coefficients equals the squared norm). Now you can conclude because $\|y_n\|$ is bounded.
