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Theorem (Approximation of compact operators) Let $X,Y$ be Banach spaces and $F: X \supset M \to Y$, where $M \ne \emptyset$, be compact. Then for every $n \in \mathbb{N}$ there exists a continuous $F_n: M \to Y$ such that $\max_{v \in M} \| F_n v - F v \| \le \frac{1}{n}$.

Proof. Let $n \in \mathbb{N}$ and $(v_k)_{k = 1}^{m}$ be a finite $\frac{1}{n}$-net for $F(M)$. Then we have $$ \min_{j = 1}^{m} \| F u - v_j \| \le \frac{1}{n} \qquad \forall u \in M. $$ For $u \in M$, $n \in \mathbb{N}$ and $j \in \{1, \ldots, m\}$ let \begin{equation} a_j(u) := \max\left(0, \frac{1}{n} - \| F u - v_j \|\right) \ge 0, \qquad F_n u := \frac{\sum_{j = 1}^{m} a_j(u) v_j}{\sum_{j = 1}^{m} a_j(u)}, \end{equation} which are well-defined and continuous, as is $u \mapsto \| F u - v_j \|$.

For $v \in M$ we have \begin{align} \| F_n v - F v \| & = \left\| \left( \sum_{j = 1}^{m} a_j(v) \right)^{-1} \sum_{j = 1}^{m} a_j(v) v_j - F v \right\| \\ & = \left\| \left( \sum_{j = 1}^{m} a_j(v) \right)^{-1} \sum_{j = 1}^{m} a_j(v) (v_j - F v) \right\| \\ & \le \left( \sum_{j = 1}^{m} a_j(v) \right)^{-1} \left( \sum_{j = 1}^{m} a_j(v) \right) \| v_j - F v \| < \frac{1}{n} \end{align} My question In the last line there is a $v_j$ outside of any summations over $j$, so there must be at least a typo. But I still don't understand, we have that $\min_{j = 1}^{m} \| F v - v_j \| \le \frac{1}{n}$, so how can $\| v_j - F v \|< \frac{1}{n}$ for any $j \in \{1, \ldots, m\}$?

Here's a shorter formulation of the proof, which unfortunately doesn't clear up my misunderstanding.

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  • $\begingroup$ If $\|Fv-v_j\|\ge 1/n$ then $\alpha_j(v)=0$. $\endgroup$ – Jochen Sep 26 at 13:37
  • $\begingroup$ @Jochen right. But what about the typo mentioned. What does actually belong there? $\endgroup$ – Viktor Glombik Sep 26 at 14:22
  • $\begingroup$ The closing parenthesis is missplaced. $\endgroup$ – Jochen Sep 26 at 16:17
  • $\begingroup$ Or the term in question has to be replaced by $\max_{j=1,..,m}\min(1/n,\|v_j-Fv\|)$, as individually $a_j(v)\|v_j-Fv\|=a_j(v)\min(1/n,\|v_j-Fv\|)\le a_j(v)/n$. Or just simply put the bound $1/n$ there. $\endgroup$ – Dr. Lutz Lehmann Sep 26 at 17:58
  • $\begingroup$ That depends on how much you want to argue. To get equality you would need that for any index $j$ with $\|Fu-v_j\|<\frac1n$ so that $a_j(u)>0$ you get $\|Fu-v_j\|=\frac1n$, which of course is absurd. However, it seems to be not as easy to find a general bound. While $ a_j(u)\|Fu-v_j\| = a_j(u)(\tfrac1n-a_j(u)) \le\frac1{4n^2}$ I do not see how to extract a bound on the second factor that is smaller than $\frac1n$. $\endgroup$ – Dr. Lutz Lehmann Sep 27 at 10:18
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One variant to repair this and to obtain the strict inequality directly is to pick an index $j^*$ with $\|Fu-v_{j^*}\|=\min_j\|F_u-v_j\|<\frac1n$. Then single this index out in the summation \begin{align} \|Fu-F_nu\|\le... &\le\left(\sum_{j:a_j(u)>0}a_j(u)\right)^{-1}\left(\sum_{j:a_j(u)>0}a_j(u)\|Fu-v_j\|\right)\\ &\le\left(\sum_{j:a_j(u)>0}a_j(u)\right)^{-1}\left(a_{j^*}(u)\|Fu-v_{j^*}\|+\frac1n\sum_{j\ne j^*:a_j(u)>0}a_j(u)\right)\\ &=\frac1n-\left(\sum_{j:a_j(u)>0}a_j(u)\right)^{-1}a_{j^*}(u)\left(\frac1n-\|Fu-v_{j^*}\|\right) <\frac1n \end{align}

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