Unit ball of $L^2$ is not compact

For \(n\ge1\), define \[ e_n(x)=\sqrt2\sin(n\pi x),\qquad 0\le x\le1. \] The functions \((e_n)\) are orthonormal in \(L^2([0,1])\), so \[ \|e_n\|_2=1 \] for every \(n\). Thus every \(e_n\) lies in the closed unit ball.

If \(m\ne n\), orthogonality gives \[ \|e_n-e_m\|_2^2 =\|e_n\|_2^2+\|e_m\|_2^2-2\operatorname{Re}\langle e_n,e_m\rangle =2. \] Hence \[ \|e_n-e_m\|_2=\sqrt2 \qquad(m\ne n). \] Therefore the sequence \((e_n)\) has no Cauchy subsequence, and hence no norm-convergent subsequence. A compact metric space is sequentially compact, so the closed unit ball of \(L^2([0,1])\) is not compact.