Linear operator not bounded is not compact

Question

Let $B$ a banach space infinite dimensional, let X a normed space, and $T: B \to X$ a linear operator such that $\|T(x)\|_{X}\geq c \|x\|_{B}$ for all $x\in B$ and $c>0$ Then $T$ is not compact

This theorem perhaps extends the fact that a continuous function sends compacts into compacts and since $T$ is not bounded it is therefore not continuous

Can you help me?

The same question has been answered here https://math.stackexchange.com/questions/3732021/let-f-be-a-bounded-linear-operator-x-to-x-s-t-that-fx-geq-m-x-fo?rq=1 — Ryszard Szwarc, Mar 19 '22 at 17:44

score 0 · Answer 1 · answered Mar 19 '22 at 05:00

0

The inequality you have stated is very different from the statement that $T$ is not bounded.

Suppose $\|x_n\| \leq 1$ for al $n$. If $T$ is compact then there is a subsequence $(Tx_{n_k})$ which is convergent, hence Cauchy. The given inequality shows that $(x_{n_k})$ is Cauchy. ($\|x_{n_k}-x_{n_j}\|\leq \frac 1 c \|T(x_{n_k})-T(x_{n_j})\| \to 0$). Since $B$ is a Banach space it follows that this sequence is convergent. We have proved that the closed unit ball of $B$ is (sequentially) compact and this implies that $B$ is finite dimensional.

answered Mar 19 '22 at 05:00

Kavi Rama Murthy

311,013

when do you proof that $T$ is NOT compact?? – wessi Mar 20 '22 at 00:07
Assuming that $T$ is compact I got a contradiction that $B$ is finite dimensional. So $T$ cannot be compact. @MathisAwesome – Kavi Rama Murthy Mar 20 '22 at 00:09

Linear operator not bounded is not compact

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