To check if operator $T:L^2[0,1]\to L^2[0,1]$ defined by $(Tf)(t)=tf(t)$ is compact or not.

I have to check if operator $$T:L^2[0,1]\to L^2[0,1]$$ defined by $$(Tf)(t)=tf(t)$$ is compact or not. The hint given is consider sequence of functions $$f_n=\sin(2\pi nt)$$. So basically I thought that this will be a bounded sequence in $$L^2[0,1]$$ such that sequence $$(Tf_n)$$ will not have any convergent subsequence. This will say $$T$$ is not compact. So we get $$\|f_n\|_2=\sqrt{\frac{1}{2}}$$, $$\forall n.$$ So this is indeed a bounded sequence. I have problem showing why it won't admit convergent subsequence. Let's say $$g_n=Tf_n$$. Then I considered for $$n \ne m$$, \begin{align}\int_{0}^{1}\lvert(g_n -g_m)(t)\rvert^2 dt&=\int_{0}^{1}t^2\sin^2(2\pi nt)dt-\int_{0}^{1}2t^2\sin(2\pi nt)\sin(2\pi mt)dt \\[0.3cm]&\ \ \ \ \ \ + \int_{0}^{1}t^2\sin^2(2\pi mt) dt. \end{align} The first integral evaluates to $$\frac{1}{2}-\frac{1}{3}-\frac{1}{(4\pi n)^2}$$. The third integral evaluates to $$\frac{1}{2}-\frac{1}{3}-\frac{1}{(4\pi m)^2}$$. The second integral evaluates to $$\frac{1}{2\pi^2(n-m)^2}-\frac{1}{2\pi^2(n+m)^2}$$. But from here I can't see why $$g_n$$ will not admit a convergent subsequence.

• So what is the sum? It looks to me like it is $1/3$ plus some stuff that goes to zero when $n$ and $m$ both go to infinity. Thus $\| g_n - g_m \|$ is always bounded below for any $n,m$, so there can't be a Cauchy subsequence. (This is assuming your computation is correct.)
– Ian
Commented Jun 1, 2020 at 19:50
• @Ian I also thought the same but the term $\frac{1}{2\pi ^2 (n-m)^2}$ is creating difficulty. That's why I am bit confused Commented Jun 1, 2020 at 19:52
• If it would be Cauchy, then in particular for $m=2n$ this bound should work. Does it indeed work in your case? Commented Jun 1, 2020 at 19:57
• @Believer $\frac{1}{2\pi^2(n-m)^2}$ is indeed not going to zero automatically but that doesn't hurt anything because this term is positive. The negative terms are all eventually less than $1/3$ in total.
– Ian
Commented Jun 1, 2020 at 20:04

So your integral is of the form $$\tfrac13+h(m,n),$$ where $$h(m,n)\to0$$ when $$m,n\to\infty$$. So choose $$m_0,n_0$$ such that $$|h(m,n)|<\tfrac16$$ for all $$m\geq m_0$$, $$n\geq n_0$$, and then the sequence $$\{g_n\}_{n\geq\max\{m_0.n_0\}}$$ admits no convergent subsequence.
Now, for a much easier argument, the spectrum of $$T$$ is $$[0,1]$$. It has nonzero accumulation points (or, it's not countable) and so $$T$$ is not compact.