Let $A_N: L^2([0,2\pi])\to L^2([0,2\pi])$ be given by

$$ f(x) \mapsto (A_Nf)(x) = \sum_{n=-N}^N \hat f(n) e^{inx} $$ where $$ \hat f(n) = \frac{1}{2\pi}\int_0^{2\pi} f(x) e^{-inx} dx $$

For, say $N=0$, we have

$$ (A_0f)(x) = \hat f(0), $$ which is just a constant, i.e. the image has dimension one. So $A_0$ is a finite rank operator and so it is compact. And similarly for any value of $N$ we take, $A_N$ will be a finite rank operator and hence compact.

Now define the operator $A: L^2([0,2\pi])\to L^2([0,2\pi])$ by

$$ f(x) \mapsto (Af)(x) = \sum_{n=-\infty}^\infty \hat f(n) e^{inx}. $$ This operator is clearly the limit of the sequence of finite rank operators $\{A_N\}$ so it is compact. However, it is also just the identity operator as it simply gives the Fourier decomposition of $f$. I.e. $Af = If = f$. But it holds that the identity operator, and hence $A$, is not compact in an infinite dimensional space.

So where have I gone wrong? Is $A$ compact or not?

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    $\begingroup$ It's not the limit of $A_N$ in the norm topology, only in the strong (and hence also in the weak) operator topology. $\endgroup$ Commented Apr 3, 2018 at 19:46

1 Answer 1


The theorem that the limit of a sequence of finite-rank operators (or compact operators) is compact refers to the norm topology on the space of continuous linear maps. If $(T_n)$ is a sequence of compact operators in $L(X,Y)$ (where $Y$ is a Banach space, $X$ a normed space) and $T_n \to T$ in the norm topology of $L(X,Y)$, then $T$ is compact.

The sequence $(A_N)$ is not norm-convergent. We have $A_N(f) \to f$ in the norm topology of $L^2([0,2\pi])$, which means that $A_N \to \operatorname{id}$ in the strong operator topology, but that is a much weaker topology than the norm topology on $\mathscr{B}(L^2([0,2\pi]))$.

  • $\begingroup$ That's great, I get it now cheers! $\endgroup$ Commented Apr 17, 2018 at 9:09
  • $\begingroup$ You say that $(A_n)$ is not norm-convergent in $\mathcal{B}(L^2([0,2\pi]))$, i.e. there exists no element $A\in \mathcal{B}(L^2([0,2\pi]))$ such that $\lim_{N\to \infty} ||A_N - A|| = 0$. Have you got any tips on how I can prove that? Is there a general procedure I can use to show it? $\endgroup$ Commented Apr 17, 2018 at 9:32
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    $\begingroup$ Since we have $A_N(f) \to f$ for every $f \in L^2$, the only candidate for a limit of (any subsequence of) $(A_N)$ is the identity. So we need to see that $\lVert A_N - \operatorname{id}\rVert$ remains large (or, if we only want to consider convergence [or not] of the full sequence, that norm needs to be large infinitely often, not necessarily always). Where "large" means there is an $\varepsilon > 0$ such that the norm is $\geqslant \varepsilon$. For the particular $(A_N)$ we have here, that's easy, taking $e^{inx}$ for an $n$ with $\lvert n\rvert > N$ shows that the norm is at least $1$. $\endgroup$ Commented Apr 17, 2018 at 9:39
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    $\begingroup$ The general strategy is similar. If a sequence converges in the norm topology, it also converges - to the same limit - in weaker topologies, like the strong or weak operator topology. That way you can often more easily find the only candidate for the limit. Once you have the candidate, it's usually not too difficult to find whether the sequence converges in the norm topology or only in weaker topologies. $\endgroup$ Commented Apr 17, 2018 at 9:43
  • $\begingroup$ Excellent, thanks for clarifying! $\endgroup$ Commented Apr 17, 2018 at 14:47

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