Let $(X,\lVert\cdot\Vert_x)$ and $(Y,\lVert\cdot\Vert_y)$ be normed spaces, X be infinite dimensional and $T\in\mathcal{L}(X,Y)$ Which has the property: there exists $m>0$ such that $ \Vert{T x}\rvert|_Y \ge m\Vert{x}\Vert_X$ for all $x \in X $. Prove that $T$ cannot be a compact operator.
I tried to solve it. I have a idea but Im not sure. If $T$ is compact if and only if $T$ is continuous and It have to be closed and bounded.
I hope someone can help me.
If a normed space $X$ is infinite dimensional, then the closed unit ball is not compact, neither precompact, which means that there exists a sequence $\{x_n\}\subset X$, without a Cauchy subsequence. In your case, as $\|Tx_i-Tx_j\|_Y\ge m\|x_i-x_j\|$, then the sequence $Tx_n$ does not have a Cauchy subsequence, and this implies that $T$ is not a compact operator.