# 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

$T:X\longrightarrow Y$

 $Tx=\lim_{n\rightarrow\infty}T_{n}x$

is linear and . Moreover, the sequence $(\|T_{n}\|)$ is bounded (http://planetmath.org/Bounded).

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

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

Then for each $x\in X$ we have

 $\|Tx\|=\lim_{n\rightarrow\infty}\|T_{n}x\|\leq M\|x\|$

which means that $T$ is . $\square$

Title pointwise limit of bounded operators is bounded PointwiseLimitOfBoundedOperatorsIsBounded 2013-03-22 17:32:10 2013-03-22 17:32:10 asteroid (17536) asteroid (17536) 4 asteroid (17536) Corollary msc 46B99 msc 47A05