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.

  • $\begingroup$ 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.) $\endgroup$
    – Ian
    Commented Jun 1, 2020 at 19:50
  • $\begingroup$ @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 $\endgroup$
    – ogirkar
    Commented Jun 1, 2020 at 19:52
  • $\begingroup$ If it would be Cauchy, then in particular for $m=2n$ this bound should work. Does it indeed work in your case? $\endgroup$ Commented Jun 1, 2020 at 19:57
  • 1
    $\begingroup$ @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. $\endgroup$
    – Ian
    Commented Jun 1, 2020 at 20:04

1 Answer 1


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.


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