show that for every $y$ the sequence $(\|\phi_{x_{n}}(y) \|)_{n \in \mathbb{N}}$ is bounded. 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?
 A: 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). 
