# Question on step in proof of approximation theorem for compact operators

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 $$$$a_j(u) := \max\left(0, \frac{1}{n} - \| F u - v_j \|\right) \ge 0, \qquad F_n u := \frac{\sum_{j = 1}^{m} a_j(u) v_j}{\sum_{j = 1}^{m} a_j(u)},$$$$ which are well-defined and continuous, as is $$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.

• If $\|Fv-v_j\|\ge 1/n$ then $\alpha_j(v)=0$. – Jochen Sep 26 at 13:37
• @Jochen right. But what about the typo mentioned. What does actually belong there? – Viktor Glombik Sep 26 at 14:22
• The closing parenthesis is missplaced. – Jochen Sep 26 at 16:17
• 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. – Dr. Lutz Lehmann Sep 26 at 17:58
• 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 $\frac1n$. – Dr. Lutz Lehmann Sep 27 at 10:18

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}