This is Exercise 2.17.1 in What is a Quantum Field Theory by Michel Talagrand. In the notation, inner products are antilinear in the first argument and the complex conjugate is denoted by an asterisk.
Consider a separable Hilbert space with an orthonormal basis $(e_n)_{n\ge0}$. [... The] operators $a$ and $a^\dagger$ [are] defined on the domain $$\mathcal{D}=\left\{\sum\limits_{n\ge0}\alpha_ne_n:\sum\limits_{n\ge0}n|\alpha_n|^2<\infty\right\}$$ by $$a(e_n)=\sqrt ne_{n-1}\ ;\quad a^\dagger(e_n)=\sqrt{n+1}e_{n+1}.$$ Exercise Prove that $a^\dagger$ is the adjoint of $a$. Prove in particular that if $|(y,a(x))|\le C\Vert x\Vert$ for $x\in\mathcal{D}$ then $y\in\mathcal{D}$.
This boils down to: given
- $ y=\sum\limits_{n\ge0}y_ne_n$ such that $\Vert y\Vert^2=\sum\limits_{n\ge0}|y_n|^2<\infty$
- $C > 0$
- For all $ x=\sum\limits_{n\ge0}x_ne_n$, if $\sum\limits_{n\ge0}n|x_n|^2<\infty$ then $\sum\limits_{n\ge0}{y_n}^{\!\ast} x_{n+1}\sqrt{n+1}\le C\sqrt{\sum\limits_{n\ge0}|x_n|^2}$
prove that $\sum\limits_{n\ge0}n|y_n|^2<\infty$.
My thoughts: let $x=\sum\limits_{n\ge0}x_ne_n$ where $x_0=0$ and $x_n=y_{n-1}/\sqrt n$ for $n\ge1$. Then $a(x)=y$ so $(y, a(x))=\Vert y\Vert^2$, and $$\sum\limits_{n\ge0}n|x_n|^2=\sum\limits_{n\ge1}n\left|\frac{y_{n-1}}{\sqrt n}\right|^2=\sum\limits_{n\ge1}|y_{n-1}|^2=\Vert y\Vert^2<\infty$$ so $x\in\mathcal{D}$, and $$\Vert x\Vert=\sqrt{\sum\limits_{n\ge0}|x_n|^2}=\sqrt{\sum\limits_{n\ge1}\left|\frac{y_{n-1}}{\sqrt n}\right|^2}=\sqrt{\sum\limits_{n\ge0}\frac{|y_n|^2}{n+1}}.$$ By our assumption that $|(y,a(x))|\le C\Vert x\Vert$ we have $\Vert y\Vert^2\le C\sqrt{\sum\limits_{n\ge0}\frac{|y_n|^2}{n+1}}$ or $\displaystyle\frac1{\sqrt{\sum\limits_{m\ge0}\frac{|y_n|^2}{n+1}}}\le\frac{C}{\Vert y\Vert^2}$.
Now I'm in a hole! I can try to continue by computing the reciprocal square root of the series. Was this a bad choice of $x$ (or $x$s)? Can you supply a better idea?
It turns out that Talagrand has posted solutions to selected exercises on his website. The key point of his solution to this one is to choose a family of $x$s. The choice $x=\sum\limits_{n=0}^k\frac{n}{\sqrt{n+1}}y_n e_{n+1}$, for arbitrary $k\ge 1$, works out well. Clearly $x\in\mathcal{D}$ since the sum is finite, so $$\sum\limits_{n=0}^kn|y_n|^2=|(y,a(x))|\le C\Vert x\Vert\le C\sqrt{\sum\limits_{n=0}^kn|y_n|^2},$$ whence $\sum\limits_{n=0}^k n|y_n|^2\le C^2$ and by monotone convergence $\sum\limits_{n\ge0}n|y_n|^2\le C^2<\infty$.