I have a question. If $X$ and $Y$ are Banach spaces, we have to prove that a compact linear operator is completely continuous. A mapping $T \colon X \to Y$ is called completely continuous, if it maps a weakly convergent sequence in $X$ to a strongly convergent sequence in $Y$ , i.e., $x_n\underset{n\to +\infty}\rightharpoonup x$ implies $\lVert Tx_n- Tx\rVert_Y\to 0$.


Since it's a homework question I will just give some steps.

  1. By linearity, we can assume that $x=0$.
  2. We have to show that for each subsequence of $\{Tx_n\}$, we can extract a further subsequence which converges to $0$ in norm in $Y$.
  3. A weakly converging sequence is bounded.
  4. $T$ maps bounded sets to sets with a compact closure.

Once the second steps is shown, we can conclude. Indeed, assume that $Tx_n$ doesn't converge to $0$. Then we are able to find $\delta>0$ and $A$ an infinite subset of the natural numbers such that $\lVert Tx_k\rVert_Y\geq\delta$ for each element of $A$. We can consider it as a subsequence, and we can't extract a further subsequence which converges to $0$, a contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.