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Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T \in \mathcal{L}(H,H)$ given by \begin{equation*} T\ f(x)=x \ f(x) \ \ \ \ f \in H,\ x \in[0,1] \end{equation*} Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.

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You can easily check that $T$ has no eigenvalues. However, $\lambda=\frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for $$\left( \frac 1 2 - x\right) f(x) = 1$$ and see that no solution exists. But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.

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It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.

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Let $$s = \sup_{\|f\|_2 \leq 1} \,\langle Tf,\,f\rangle=1.$$

It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $\|f_n\| \leq 1$, and $\langle Tf_n ,\, f_n \rangle \rightarrow 1$.

Assume that there is some converging subsequence $Tf_{\varphi(n)} \rightarrow g$. Then you easily get $\langle g ,\,f_{\varphi(n)}\rangle \rightarrow 1$.

We know there is a weakly converging subsequence $f_{\varphi(\psi(n))} \rightarrow h$ thus $\|g\| \leq 1$ and $\|h\| \leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.

Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.

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