# Limit of sequence of finite rank operators given by $(A_Nf)(x) = \sum_{n=-N}^N \hat f(n) e^{inx}$ is the identity operator, so how can it be compact?

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?

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

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 $\mathscr{B}(L^2([0,2\pi]))$.
• 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? Commented Apr 17, 2018 at 9:32
• 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 $1$. Commented Apr 17, 2018 at 9:39