Let $X$ be a Banach space and $B$ be the closed unit ball contained in $X$. Let $\{T_{\alpha}\}$ be a family of bounded linear operators from $B$ to $V$, a normed vector space. My question is: Suppose the $T_{\alpha}$ are equicontinuous, why must the $T_{\alpha}$ be pointwise bounded?
My work so far: In the definition of equicontinuity, take $\varepsilon = 1$, then the definition of equicontinuity gives a resulting $\delta$. Cover $B$ by balls of radius $\delta$. Since $B$ is compact, there exists finitely many points $\{x_{i}\}_{i = 1}^{N}$ such that $X \subset \bigcup_{i = 1}^{N}B_{\delta}(x_{i})$. Fix an $x \in B$. I want to show that $\{T_{\alpha}x\}$ is bounded with the bound independent of $\alpha$. Since $x \in B$, there exists an $x_{j}$ such that $\|x - x_{j}\| < \delta$. Then $\|T_{\alpha}x - T_{\alpha}x_{j}\| \leq 1$ and hence $\|T_{\alpha}x\| \leq 1 + \|T_{\alpha}x_{j}\|$. However, my bound on the left hand side could depend on $\alpha$. How would I get rid of this?
The problem is that the closed unit ball $B$ is not compact when $X$ is infinite dimensional.
However, using the definition of equi-continuity, we get $\delta$ such that if $\lVert x-y\rVert\lt \delta$, then for each $\alpha$, $\lVert T(x)-T(y)\rVert\leqslant 1$. In particular, with $y=0$, we get $\lVert T_\alpha(x)\rVert\leqslant 1$ for each $\alpha$ if $\lVert x\rVert\lt \delta$.
If $u\neq 0$, then $\left\lVert \frac{\delta}{2\lVert u\rVert}u\right\rVert\lt\delta$, hence $\lVert T_\alpha(u)\rVert\leqslant \frac 2\delta \lVert u\rVert$.