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.

  • $\begingroup$ Assume $T$ is compact. If $(x_n)$ is bounded in $X$, then there is a subsequence $(x_{n_k})$ such that $(Tx_{n_k})$ is norm-Cauchy. Using the given "bounded below" condition, what does this say about $(x_{n_k})$? $\endgroup$ – David Mitra Jan 27 '14 at 23:15
  • $\begingroup$ $x_{nk}$ is convergent @DavidMitra ? $\endgroup$ – Rabia K. Jan 27 '14 at 23:46
  • $\begingroup$ At this point we can at least say it is norm-Cauchy. So then, every bounded sequence, in particular every sequence in the unit ball of $X$, has a norm-Cauchy subsequence. What does this tell you? $\endgroup$ – David Mitra Jan 27 '14 at 23:48
  • $\begingroup$ Banach space that is weakly complete as well as norm compute so every bounded sequence in $X$ has a convergent subsequence .This means the unit ball of $X$ norm compact.@DavidMitra $\endgroup$ – Rabia K. Jan 28 '14 at 0:08
  • $\begingroup$ The only fact you need is that a normed space has a relatively compact unit ball if and only if it is finite dimensional (in fact if $X$ is infinite dimensional, there is a sequence $(x_n)$ of norm one vectors such that $\Vert x_n-x_m\Vert>1/2$ whenever $n\ne m$). So if $T$ is compact, then $X$ must be finite dimensional. $\endgroup$ – David Mitra Jan 28 '14 at 0:11

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.

  • $\begingroup$ Thank you for your answer @YiorgosS.Smyrlis $\endgroup$ – Rabia K. Jan 28 '14 at 10:53

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