Separable Hilbert Space Let $\mathcal{H}$ be a separable Hilbert space with complete orthonormal system $e_{1},e_{2},\ldots$
(a) Let $T:\mathcal{H}\rightarrow\mathcal{H}$ be a continuous linear operator. Suppose that there is a sequence $(v_{i})_{i\geq1}$ of strictly positive numbers, and numbers $\lambda,\mu\geq 0$, with the property that,
\begin{align}
\sum_{j=1}^{\infty} |(Te_{j}|e_{i})|v_{j} &\leq\lambda v_{i}\\[3pt]
\sum_{i=1}^{\infty} |(Te_{j}|e_{i})|v_{i} &\leq\mu v_{j},
\end{align}
for all $i,j\geq 1$. Show that $\|T\|\leq\sqrt{\lambda\mu}$.
Hint: Apply Parseval's Identity to $\|Tx\|^{2}$. Expand $x$ into it's Fourier series to see that
\begin{align}
|(Tx|e_{i})|\leq\sum_{j=1}^{\infty}|(x|e_{j})||(Te_{j}|e_{i})|.
\end{align}
Then write $|(x|e_{j})||(Te_{j}|e_{i})|=\bigg(|(x|e_{j})|v_{j}^{-1/2}|(Te_{j}|e_{i})|^{1/2}\bigg)\bigg(|(Te_{j}|e_{i})|^{1/2}v_{j}^{1/2}\bigg)$.
(b) Suppose that $T:\mathcal{H}\rightarrow\mathcal{H}$ is a continuous linear operator with,
\begin{align}
(Te_{j}|e_{i})=(i+j-1)^{-1}.
\end{align}
Show that $\|T\|\leq\pi$.
Hint: Use (a) with $v_{j}=(j-\frac{1}{2})^{-1/2}$, and approximate sums by integrals.
I have completed part (a), but I am confused by part (b) of the question. When the hint mentions to approximate sums by integrals, is it talking about the inequalities given in part (a)? or part of the working for part (a) where the inequality in the hint comes in to play?
 A: Starting from the hint one obtains
$$ \sum_{j = 1}^\infty \vert (Te_j, e_i) \vert v_j = \sum_{j = 1}^\infty \frac{1}{(i + j - 1) \sqrt{j - 1/2}}.$$
Now the summands behave like $a_j := ((i + j - 1) \sqrt{j - 1/2})^{-1} = \mathcal O(j^{-3/2})$ for each $i \in \mathbb N$ , so the sum converges for all $i \in \mathbb N$. Moreover, the sequence $(a_j)_{j \in \mathbb N}$ is decreasing for each $i \in \mathbb N$. So by the integral test  the integral 
$$ \int_1^\infty \frac{1}{(i + x - 1) \sqrt{x - 1/2}} \, dx$$ is finite for all $i \in \mathbb N$ and one has 
\begin{align*}
\sum_{j = 1}^\infty \frac{1}{(i + j - 1) \sqrt{j - 1/2}} &\leq \frac{\sqrt 2}{i} + \int_1^\infty \frac{1}{(i + x - 1) \sqrt{x - 1/2}} \, dx \\
&= \frac{\sqrt 2}{i} + 2 \frac{\arctan(\sqrt{2 i - 1})}{\sqrt{i - 1/2}} \\
&= \left(\frac{\sqrt 2 \sqrt{i - 1/2}}{i} + 2 \arctan(\sqrt{2 i - 1}) \right) v_i.
\end{align*}
Now the first factor in the estimate still depends on $i$. But the function 
$$ g : x \mapsto \frac{\sqrt 2 \sqrt{x - 1/2}}{x} + 2 \arctan(\sqrt{2 x - 1})$$
is increasing on $[1/2, \infty)$ and one clearly has that $\lim_{x \to \infty} g(x) = \pi$. So finally we obtain the estimate
\begin{align*}
\sum_{j = 1}^\infty \vert (Te_j, e_i) \vert v_j &= \sum_{j = 1}^\infty \frac{1}{(i + j - 1) \sqrt{j - 1/2}} \\
&\leq \left(\frac{\sqrt 2 \sqrt{i - 1/2}}{i} + 2 \arctan(\sqrt{2 i - 1}) \right) v_i \leq \pi v_i.
\end{align*}
By interchanging the roles of $i$ and $j$ one obtains analogously that
$$\sum_{i = 1}^\infty \vert (Te_j, e_i) \vert v_j \leq \pi v_j. $$
So (a) yields that $\Vert T \Vert \leq \sqrt{\pi^2} = \pi$ which is the desired estimate. I hope it got clear :)
