# Let $H$ be a Hilbert space and $(u_n) \subseteq H$ an orthogonal sequence. Show the functional $T(x) = \sum_n \langle x , u_n \rangle$ is bounded.

Let $$H$$ be a Hilbert space and $$(u_n)_{n \in \mathbb{N}}$$ an orthogonal sequence in $$H$$. Consider the linear functional defined by $$T(x) = \sum_{n=1}^\infty \langle x , u_n \rangle , \forall x \in H$$. Show that $$T$$ is bounded, that is $$||T|| < \infty$$.

My approach is as follows. Let $$T_n = \sum_{k=1}^n \langle x , u_n \rangle, \forall n \in \mathbb{N}$$.

I would then like to prove this using the uniform boundedness principle:

Let $$\{T_n\}_{n \in \mathbb{N}}$$ be a set bounded linear functionals. Suppose that for every $$x \in H$$, there exists a constant $$M_x$$ such that $$||T_n(x)|| \le M_x, \forall n \in \mathbb{N}$$. Then there exists a constant $$M \ge 0$$ such that $$||T_n|| \le M, \forall n \in \mathbb{N}$$.

I am able to show the required condition for the Uniform Boundedness Principle, however, I can not seem to show that each $$T_n$$ is bounded. That is, I still need to show that for each $$n \in \mathbb{N}, \exists M_n \ge 0$$ such that $$||T_n(x)|| \le M_n ||x||, \forall x \in H$$

Any help is appreciated!

EDIT: I am able to show that each $$T_n$$ is bounded, but am not able to extend this to show that $$T$$ is bounded as well. Additionally, we assume that $$\{||u_n||\}$$ is bounded.

The boundedness of $$T_n$$ follows from $$|T_n x| \le \sum_{k=1}^n \|x\| \, \|u_n\|$$ with $$M_n = \sum_{k=1}^n \|u_n\|$$.

Does your definition of orthogonal sequence includes the boundedness of $$\|u_n\|$$? Otherwise, $$T$$ is not bounded since $$\frac{T (u_n)}{\|u_n\|} = \|u_n\|.$$

• Ok. This is typically called a orthonormal system. – gerw Mar 29 '19 at 15:15
• Yes, in the larger question, we can prove that $||u_n||$ is bounded. But from there how can I proceed to show that $T$ is bounded. – Bojack Horseman Mar 29 '19 at 15:16
• I have suddenly realized that boundedness of $\|u_n\|$ is not enough: Consider $H = \ell^2$, $u_n = e_n$ (unit sequence) and $x = (1/i)_{i\in\mathbb N}$. Then, $Tx = \sum_{i=1}^\infty 1/i$ is not well defined... – gerw Mar 29 '19 at 15:18
• Ouch, unfortunate. I have not been able to prove anything stronger on the $||u_n||$s. – Bojack Horseman Mar 29 '19 at 15:20
• Can you show your proof for the condition of the uniform boundedness principle? – gerw Mar 29 '19 at 15:23

Your $$T$$ is not even well-defined. Counter-example:

Let $$\mathcal{H}=l^{2}(\mathbb{N})=\{x\in\mathbb{R}^{\mathbb{N}}\mid\sum_{n=1}^{\infty}|x(n)|^{2}<\infty\}$$ with the usual inner-product: $$\langle x,y\rangle=\sum_{n=1}^{\infty}x(n)y(n).$$ It is well-known that $$\mathcal{H}$$ is a Hilbert space.

For each $$n\in\mathbb{N}$$, let $$e_{n}\in\mathcal{H}$$ be defined by $$e_{n}(k)=\begin{cases} 1, & \mbox{if }k=n\\ 0, & \mbox{if }k\neq n \end{cases}.$$ Then $$\{e_{n}\mid n\in\mathbb{N}\}$$ is an orthonormal family (in fact, it is also total but we do not need this fact). Note that the map $$T:\mathcal{H}\rightarrow\mathbb{R}$$, $$T(x)=\sum_{n=1}^{\infty}\langle x,e_{n}\rangle$$ is not well-defined. For example, let $$x\in\mathbb{R}^{\mathbb{N}}$$be defined by $$x(n)=\frac{1}{n}$$. Note that $$\sum_{n=1}^{\infty}|x(n)|^{2}=\sum_{n=1}^{\infty}\frac{1}{n^{2}}<\infty$$, so $$x\in\mathcal{H}$$. However, $$\sum_{n=1}^{\infty}\langle x,e_{n}\rangle=\sum_{n=1}^{\infty}\frac{1}{n}$$, which is divergent.