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 $$\frac{T (u_n)}{\|u_n\|} = \|u_n\|.$$

  • $\begingroup$ Ok. This is typically called a orthonormal system. $\endgroup$ – gerw Mar 29 '19 at 15:15
  • $\begingroup$ 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. $\endgroup$ – Bojack Horseman Mar 29 '19 at 15:16
  • 1
    $\begingroup$ 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... $\endgroup$ – gerw Mar 29 '19 at 15:18
  • $\begingroup$ Ouch, unfortunate. I have not been able to prove anything stronger on the $||u_n||$s. $\endgroup$ – Bojack Horseman Mar 29 '19 at 15:20
  • $\begingroup$ Can you show your proof for the condition of the uniform boundedness principle? $\endgroup$ – 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.


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