Example of a bounded linear operator on $L^2([0,1])$ that is not compact

Question

The exercise says to prove or to give counter example whether every bounded linear functional from $L^2([0,1])$ to itself is compact. I feel that it is not true as there can be orthonormal sequence that converges weakly to zero in $L^2([0,1])$, but under identity operator does not map to strong convergent sequence. So it is not compact. It will be very helpful if anyone can give any insight.

Answer 1

The identity operator in an infinite-dimensional Banach space is never compact, because the closed unit ball in such a space is not compact. For example, you can construct a sequence $x_n$ with $\|x_n\| = 1$ and $\|x_n - x_m\| \ge 1/2$ for all $n \ne m$.