Compact operator on $L_2([0,1],m)$

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.

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.

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.

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.