1
$\begingroup$
  • Let $X$ be a Banach Space ,$Y$ a normed space and $T_n\in B(X,Y)$ such that $(T_nx)$ is Cauchy in $Y$ for every $x\in X$ .Show that $(||T_n||)$ is bounded.
  • If in addition $Y$ is complete show that $T_nx\to T(x) ;T\in B(X,Y)$ where $B(X,Y)$ denotes the set of bounded linear operators from $X$ to $Y$.

Attempt:

  • $(T_nx)$ is Cauchy $\implies $ $(T_nx)$ is bounded for each $x$ and $X$ is Banach Space hence by Uniform Boundedness Principle we have $T_nx$ is uniformly bounded and hence so $(||T_n||)$ is bounded.
  • Unable to do it. Since $Y$ is complete and $T_nx$ is Cauchy ,so $T_n(x)\to c_x$ for each $x$. But how to show that $c_x=T(x)$.

Please help me in this case and moreover is the 1st one correct?

  • 1
    Step 1: Show that $T \colon x \mapsto \lim T_n(x)$ defines a linear map (don't worry about continuity or anything yet). Step 2: Fix an arbitrary $x\in X$, and find a bound for $\lVert T(x)\rVert$ in terms of $\lVert x\rVert$ and what you know about $\lVert T_n\rVert$. Step 3: Wrap everything up. – 2017-01-06
  • 0
    For Step 2: Let $x=a$ then $||T(a)||=\lim (||T_n||||a||)\le \lim (||T_n||)(||a||)\le M||a||$; How to wrap up now – 2017-01-06
  • 0
    Will you please help ;I am unable to proceed ? – 2017-01-06
  • 0
    You have a collection of bounded operators that is uniformly bounded at every point of the Banach space. So the collection must be uniformly bounded in operator norm. – 2017-01-06
  • 0
    What is the reference for this problem if you remember? @Learnmore – 2018-11-20

1 Answers 1

2

First we define a map $T \colon X \to Y$ by

$$T(x) := \lim_{n\to\infty} T_n(x).$$

The limit exists, since by assumption $\bigl(T_n(x)\bigr)$ is a Cauchy sequence in $Y$, and $Y$ is complete. By the uniqueness of limits in Hausdorff spaces, we have thus defined $T$ as a map.

As the next step, we show that $T$ is a linear map. That follows immediately from the linearity of the $T_n$ and general limit algebra,

$$T(x+y) = \lim_{n\to\infty} T_n(x+y) = \lim_{n\to\infty}\:\bigl(T_n(x) + T_n(y)\bigr) = \lim_{n\to\infty} T_n(x) + \lim_{n\to\infty} T_n(y) = T(x) + T(y)$$

for $x,y\in X$, and

$$T(\lambda x) = \lim_{n\to\infty} T_n(\lambda x) = \lim_{n\to\infty} \:\bigl(\lambda T_n(x)\bigr) = \lambda \lim_{n\to\infty} T_n(x) = \lambda T(x)$$

for $x\in X$ and $\lambda \in \mathbb{C}$ (or $\lambda \in \mathbb{R}$ if we're dealing with real scalars).

Finally, we show that $T\in B(X,Y)$, using the first part. By the first part, there is an $M$ such that $\lVert T_n\rVert \leqslant M$ for all $n$, and that gives

$$\lVert T(x)\rVert = \lim_{n\to\infty} \lVert T_n(x)\rVert \leqslant \limsup_{n\to\infty} \:\bigl( \lVert T_n\rVert\cdot \lVert x\rVert\bigr) \leqslant M\lVert x\rVert,$$

whence $\lVert T\rVert \leqslant M$ follows.