Prove that the operator \(T : L^2([0,1]) \to L^2([0,1])\) defined by setting \(T[f](x) = xf(x)\) is continuous and symmetric (self-adjoint).
Prove that \(T\) is not compact.
Solution
1. Prove boundedness and self-adjointness.
For \(f\in L^2([0,1])\), \[ \|Tf\|_2^2 =\int_0^1 x^2|f(x)|^2\,dx \le \int_0^1|f(x)|^2\,dx =\|f\|_2^2. \] Thus \(T\) is bounded and \(\|T\|\le1\). In fact \(\|T\|=1\) by testing on functions supported near \(1\).
For \(f,g\in L^2([0,1])\), \[ \langle Tf,g\rangle =\int_0^1 xf(x)\overline{g(x)}\,dx =\int_0^1 f(x)\overline{xg(x)}\,dx =\langle f,Tg\rangle. \] Hence \(T=T^*\).
2. Construct a bounded sequence whose image has no convergent subsequence.
For \(n\in\mathbb Z\), set \[ e_n(x):=\sqrt2\,e^{4\pi i n x}\mathbf1_{[1/2,1]}(x). \] Then \((e_n)\) is an orthonormal sequence in \(L^2([0,1])\). Hence for \(n\ne m\), \[ \|e_n-e_m\|_2^2=2. \]
Since \(x\ge1/2\) on \([1/2,1]\), \[ \begin{aligned} \|Te_n-Te_m\|_2^2 &=\int_{1/2}^1x^2|e_n(x)-e_m(x)|^2\,dx\\ &\ge \frac14\|e_n-e_m\|_2^2\\ &=\frac12. \end{aligned} \] Thus the sequence \((Te_n)\) is pairwise separated by at least \(1/\sqrt2\), so it has no Cauchy subsequence and therefore no convergent subsequence.
The bounded sequence \((e_n)\) lies in the unit ball, but its image under \(T\) is not relatively compact. Hence \(T\) is not compact.