TITLE: Uniform boundedness principle for norm convergence
QUESTION [5 upvotes]: This is from Tao's book;

Let $X$ be Banach space, let Y be normed vector space, and let $(T_n)_{n=1}^{\infty}$ be a family of continuous linear operators $T_n : X \to Y$. Then following is equivalent.
(i) (Pointwise convergence) For every $x \in X$. $T_n(x)$ converges strongly in $Y$ as $n \to \infty$.
(ii) (Pointwise convergence to a continuous limit) There exists a continuous linear map $T:X \to Y$ such that for every $x\in X$, $T_n(x)$ converges strongly in $Y$ to $T(x)$ as $n\to\infty$.
(iii) (Uniform boundedness + dense subclass convergence) The operator norms $\{||T_n|| : n = 1, 2, . . .\}$ are bounded, and for a dense set of $x$ in $X$, $T_n(x)$ converges strongly in $Y$ as $n \to \infty$

What I have trouble is $(i) \to (iii)$ and $(iii) \to (ii)$.
My proof on $(i) \to (iii)$ is like below; second statement of $(iii)$ is easy so it suffices to prove only boundedness of the set of operator norms. Since convergent sequence is bounded, $\{ T_n(x)\}_{n=1}^{\infty}$ is clearly bounded sequence. Hence there is some $N $ such that $||T_n(x)|| < N$. Since $T_n$ is linear, $||T_n(x)|| = ||T_n ||_{\mathrm{op}}||x|| < N$, and since $x$ is fixed, $||T_n||_{\mathrm{op}}$ is bounded. Is this argument reasonable? In the Tao's note, he mention Baire category theorem to prove this.But I don't know how to prove this using Baire category theorem directly.
For second question, $(iii) \to (ii)$, I have an idea that construct $T:X \to Y$ such that $T(x) =\lim_{n \to \infty} T_n(x)$ and $T(0)=0$. But I know that $T$ is linear and continuous on dense set, but don't know how to extend it to limit point of $X$. Could you give me some hint?

REPLY [2 votes]: For (i) $\implies$ (iii). For each $N=1,2,3,\ldots,$ define $F_{N}=\{ x : \sup_{n}\|T_{n}x\| \le N \}$, and show every $x$ is in some $F_{N}$.
For (iii) $\implies$ (ii). If $\|T_{n}\|\le M$ for some $M$ and all $n$,
$$
\begin{align}
\|T_{n}x-T_{n+k}x\| & \le \|T_{n}x-T_{n}y\|+\|T_{n}y-T_{n+k}y\|+\|T_{n+k}y-T_{n+k}x\| \\
     & \le 2M\|x-y\|+\|T_{n}y-T_{n+k}y\|.
\end{align}
$$