Application of Uniform Boundedness Theorem to prove an equivalence involving sequences.

Question

After state and prove the Uniform Boundedness Theorem, the Kreyszig Functional Analysis book presents the following problem:

I'm trying to solve it but I need help to finish it. What I have done (probably the easiest part) is below.

$(b)\Rightarrow(a)$ Immediate consequence of Uniform Boundedness Theorem.

$(b)\Rightarrow(c)$ For each $x\in X$, there exists $c\in\mathbb{R}$ such that $\|T_nx\|\leq c$ for all $n\in\mathbb{N}$. Hence $|g(T_nx)|\leq\|g\|\|T_nx\|\leq\|g\|c$.

Thanks.

EDIT:

$(a)\Rightarrow(b)$ See Davide Giraudo's answer below.

$(c)\Rightarrow(b)$ For each $y\in Y$, let $J(y):Y'\to\mathbb{K}$ be defined by $J(y)(f)=f(y)$ for all $f\in Y'$ and for each $f\in Y'$ consider the sequence $(H_n(f))$, where $H_n(f)=J(T_nx)(f)\;\forall n\in\mathbb{N}$. By $(c)$ we conclude that $(H_n(f))$ is bounded for all $f$ so that (by Uniform Boundedness Theorem) $\|H_n\|$ is bounded. We know that $\|J(y)\|=\|y\|$ for all $x\in Y$. Hence $\|T_nx\|=\|J(T_nx)\|=\|H_n\|$ is bounded.

Now, my question is: why is it necessary be $Y$ a Banach space?

Answer 1

Since $\lVert T_nx\rVert\leqslant \lVert T_n\rVert\cdot \lVert x\rVert\leqslant \sup_{k\geqslant 1}\lVert T_k\rVert\cdot \lVert x\rVert$, the hardest part was the consequence of the uniform boundedness principle.