Here is the question and its solution (a solution given to me by Keefer):
Let $\{x_{n}\}$ be an unbounded sequence in Hilbert $\mathcal{H}.$ Prove that there exists a vector $x \in \mathcal{H}$ such that the sequence $\{\langle x_{n}, x \rangle\}$ is unbounded.
Solution:
Fixed answer: Suppose for every $x \in \mathcal{H}$, $(x,x_n)$ is bounded. Then for the linear functionals $\phi_{x_n}$ given by $\phi_{x_n}(y) = (y,x_n)$, we have for each $y \in \mathcal{H}$, $\sup_n |\phi_{x_n}(y)| < \infty$. Then by the uniform boundedness principle $\sup_n \|\phi_{x_n}\| < \infty$. But $\|\phi_{x_n}\| = \|x_n\|$, so $\sup_n \|x_n\| < \infty$. So $x_n$ is bounded.
My questions about the solution:
1- It is not very clear for me why the linear functional $\phi_{x_{n}}$ is defined as thought and what is the new variable $y$ (I was trying to compare what is $h$ in the statement of RRT and what is $x$ there)? I am a little bit confused. I know the solution is trying to use Riesz FrechetRepresentation theorem given below:
2- I do not know why in the solution: $\sup_n |\phi_{x_n}(y)| < \infty$ is it by Riesz representation theorem because $\|T(h)\| = \|h\|$ but in the RRT we have $T$ and $T_{h}.$ why if $T$ is isometric that means that $\|T(h)\| = \|h\|$?
3-For the application of the uniform bounded principle for $\phi_{x_n}(y) = (y,x_n)$:
I know that $\phi_{x_n}$ is bounded by our assumption as we are assuming the contrapositive. but how is for every $y$ the sequence $(\|\phi_{x_{n}}(y) \|)_{n \in \mathbb{N}}$ is bounded? Is this because of the definition of the inner product by means of the norm? I do not know, could anyone explain this for me please?
Could anyone help me in answering this questions please?
This has nothing to do with the Riesz Representation Theorem (which shows that every linear functional on a Hilbert space is of that form; you don't need that here, as the functionals are constructed explicitly).
A linear functional is a functional that takes an element in $H$ and gives you a number, in a linear way. A function needs an argument, which in the answer you quote was named $y$. So, for each $n$, one is defining the function $\phi_n:y\longmapsto \langle y,x_n\rangle$. This is trivially linear: $$ \phi_n(y_1+y_2)=\langle (\alpha y_1+y_2),x_n\rangle=\alpha\langle y_1,x_n\rangle+\langle y_2,x_n\rangle=\alpha\phi_n(y_1)+\phi_n(y_2). $$ And bounded: by Cauchy-Schwarz, $$ |\phi_n(y)|=|\langle y,x_n\rangle\leq\|x_n\|\,\|y\|,\ \ \ \ \ y\in H. $$
In the answer, $\langle x,x_n\rangle$ is assumed bounded. Because one is proving the contrapositive: you want "$\{x_n\}$ unbounded implies that there exists $x$ with $\{\langle x,x_n\rangle\}$ unbounded", and its contrapositive is "if $\{\langle x,x_n\rangle\}$ is bounded for all $x$, then $\{x_n\}$ is bounded".
So, the assumption for the contrapositive is that $\sup_n\{|\langle x,x_n\rangle|\}<\infty$. That's precisely $\sup_n |\phi_n(x)|<\infty$. Then the UBP gives you that $\sup_n|\phi_n|<\infty$. And now you use that $\|\phi_n\|=\|x_n\|$ (it's an easy exercise, but you can get it from the Riesz Representation Theorem if you want).