uniform convergence on compact subsets of the linear,continuous and uniformly bounded operators.

Question

Let $X,Y$ be normed spaces. Let $T_j : X\to Y$ be a sequence of linear and continuous functions, such that $\lVert T_j\rVert\lt K$ $\forall j$. If $T_j$ converges pointwise to $T$, prove that $T$ is also linear and continuous, and the convergence is uniform on compact subsets of $X$. I proved everything except for the uniform convergence on compact subsets. Please help me )=!

Answer 1

Let $M \subseteq X$ be compact and $\varepsilon>0$. There exist $x_1,\ldots,x_m \in M$ such that

$$M \subseteq \bigcup_{j=1}^m B(x_j,\varepsilon/K)$$

Consequently, for any $x \in M$ we can choose $j \in \{1,\ldots,m\}$ such that $x \in B(x_j,\varepsilon/K)$. By the triangle inquality,

$$\begin{align*} |T_n(x)-T(x)| &=|(T_n(x)-T_n(x_j))+(T_n(x_j)-T(x_j))+(T(x_j)-T(x))| \\ &\leq \|T_n\| \cdot |x-x_j| + |T_n(x_j)-T(x_j)| + \|T\| \cdot |x-x_j| \end{align*}$$

for all $n \in \mathbb{N}$. For $n \geq n_0$ sufficiently large, we have

$$|T_n(x_j)-T(x_j)| \leq \varepsilon \qquad (j =1,\ldots,m)$$

since $T_n(x_j) \to T(x_j)$ for $j=1,\ldots,m$. From $|x-x_j| < \varepsilon/K$ and $\|T_n\| \leq K$, $\|T\| \leq K$ we conclude

$$|T_n(x)-T(x)| \leq \varepsilon+\varepsilon+\varepsilon = 3\varepsilon$$

for all $n \geq n_0$. Since $n_0$ does not depend on $x$, this finishes the proof.