# Characterization of boundedness of a linear operator under compactness

The following is an exercise from An Introduction to Banach Space Theory by Robert E. Megginson.

Suppose that $$T$$ is a linear operator from a normed space $$X$$ into a normed space $$Y$$. Prove that the following are equivalent.

a) The operator $$T$$ is continuous.

b) The set $$T(K)$$ is a compact subset of $$Y$$ whenever $$K$$ is a compact subset of $$X$$.

c) The set $$T(K)$$ is a weakly compact subset of $$Y$$ whenever $$K$$ is a weakly compact subset of $$X$$.

The idea that I have is that, to prove that b) implies a), let $$x_{n}\rightarrow 0$$, and consider the sequentially compact set $$K=\{x_{n}:n=1,2,...\}\cup\{0\}$$, then $$K$$ is also compact, and hence $$T(K)=\{T(x_{n}):n=1,2,...\}\cup\{0\}$$ is sequentially compact, it seems that we cannot say immediately that $$T(x_{n})\rightarrow 0$$, this is where I got stuck.

To prove that a) and c) are equivalent, I have only the idea that norm-to-norm continuous is equivalent to weak-to-weak continuous, but what can I do next?

The space $$X$$ has no approximate property, so one cannot use some sort of approximated by compact operators to argue this exercise.

Edit:

Part of the answer has been addressed here. That is, by Mazur compactness theorem, weakly compact is equivalent to weakly sequentially compact, so the statement c) can be rephrased as the following one.

The set $$T(K)$$ is sequentially compact subset of $$Y$$ whenever $$K$$ is a weakly sequentially compact.

To show that c) implies a). Assume that some sequence $$\{x_{n}\}$$ is such that $$x_{n}\rightarrow 0$$ but $$\|T(x_{n})\|\rightarrow\infty$$, then the set $$K=\{x_{n}:n=1,2,...\}\cup\{0\}$$ is weakly sequentially compact which implies by assumption that $$T(K)=\{T(x_{n}): n=1,2,...\}\cup\{0\}$$ is also weakly sequentially compact. In particular, the sequence $$\{T(x_{n})\}$$ has a weakly convergent sequence and hence the subsequence is bounded, this contradicts that $$\|T(x_{n})\|\rightarrow\infty$$.

In the same fashion, one proves that b) implies a).

To show that a) implies c). As I have said, $$T$$ is weak-tp-weak continuous. Assume that $$K$$ is a weakly compact subset of $$X$$ but $$T(K)$$ is not, then by sequential characterization again, one finds some sequence $$\{T(x_{n})\}$$ such that $$T(x_{n})$$ has no weakly convergent subsequence, where $$\{x_{n}\}\subseteq K$$. But $$\{x_{n}\}$$ has a weakly convergent subsequence, say, $$\{x_{n_{k}}\}$$, then by weak-to-weak continuity of $$T$$, $$\{T(x_{n_{k}})\}$$ is weakly convergent, this is a contradiction.

In the same fashion, one proves that a) implies b).

• What's the problem with $T(x_n) \to 0$? If you assume $T$ to be continous and $x_n \to 0$ then you get $T(x_n) \to 0$. – eddie Dec 28 '18 at 21:18
• Sorry, I wrote the wrong a) and b). – user284331 Dec 28 '18 at 21:19
• Anyway, it seems that I have figured out the solution. – user284331 Dec 28 '18 at 21:23
• Yes it's not necessary to show that $Tx_n \to 0$, it sufficies to show that $Tx_n$ is bounded. – eddie Dec 28 '18 at 21:32

Proof of b) implies a): if $$T$$ is not continuous there exists $$\{y_n\}$$ such that $$\|Ty_n\| >n^{2}\|y_n\|$$. Let $$x_n =\frac {y_n} {n \|y_n\|}$$. Then $$x_n \to 0$$ and $$\|T(x_n)\|>n$$. Hence $$\{0,x_1,x_2,\cdots\}$$ is a compact set whose image is not bounded.