pointwise limit of bounded operators is bounded

The following result is a corollary of the principle of uniform boundedness.

Theorem - Let $X$ be a Banach space and $Y$ a normed vector space. Let $(T_n)\in B(X,Y)$ be a sequence of bounded operators from $X$ to $Y$. If $(T_{n}x)$ converges for every $x\in X$, then the operator

$T:X⟶Y$

$$Tx=\underset{n\to \mathrm{\infty }}{lim}{T}_{n}x$$

is linear and . Moreover, the sequence $(\parallel T_n\parallel )$ is bounded (http://planetmath.org/Bounded).

Proof : It is clear that the operator $T$ is linear.

For each $x\in X$ we have since $(T_{n}x)$ is . By the principle of uniform boundedness (http://planetmath.org/BanachSteinhausTheorem) we must also have .

Then for each $x\in X$ we have

$$\parallel Tx\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel T_{n}x\parallel \le M\parallel x\parallel$$

which means that $T$ is . $\mathrm{\square }$