# Question regarding sequence of continuous functions

$$\left \{f_n \right \}$$ is a sequence of continuous non-negative functions defined on $$[0,1]$$, such that $$\lim_{n\rightarrow\infty} f_n(x) = 0$$ pointwise on $$[0,1]$$.

I am asked to prove that $$\forall \epsilon > 0$$, $$\exists \delta > 0, N \in \mathbb{N}$$ and points $$x_1, ... , x_N$$ and $$n_1, ... , n_N$$ such that:

$$[0,1] \subset \cup_{k = 1}^{N} [ x_k - \delta, x_k + \delta]$$

And that:

$$0 \leq f_{n_k} (x) < \epsilon \hspace{6pt}\forall x \in [x_k - \delta, x_k + \delta] \hspace{6pt}\forall 1 \leq k \leq N$$

Here is my solution so far:

Fix $$M \in \mathbb{N}$$ large, and let $$\epsilon > 0$$. Then each function in sequence is continuous on compact domain, so uniform continuous, so $$\forall n \in \mathbb{N}$$, there exists $$\delta_n > 0$$ such that for all $$x,\tilde{x}$$ with $$|x-\tilde{x} | < \delta_n$$ we have that $$|f_n(x) - f_n(\tilde{x})| < \epsilon / 2$$.

Let $$\delta = \min \{\delta_1, ... , \delta_M \}$$, then by total boundedness, we can find points $$x_1, ... , x_N$$ such that $$[0,1] \subset \cup_{k = 1}^{N} [ x_k - \delta, x_k + \delta]$$. Then $$\forall k \leq N$$ and $$\forall i \leq M$$, we have that $$0 \leq f_i(x) < f_i(x_k) + \epsilon / 2 \hspace{10pt} \forall x \in [x_k - \delta, x_k + \delta]$$ but since $$\lim_{n \rightarrow \infty} f_n(x) = 0$$, then there exists an $$n_k$$ such that $$f_{n_k}(x_k) < \epsilon / 2$$.

If $$n_k \leq M$$, then we can combine with inequality above to get: $$0 \leq f_{n_k} (x) < f_{n_k}(x_k) + \epsilon / 2 < \epsilon$$

But what about when $$n_k > M$$? If we increase $$M$$, then $$\delta$$ may decrease, causing $$N$$ to increase. This is what I am having trouble resolving. Any help would be very much appreciated.

• Have you thought defining a new function as the maximum of the $N$ many you intend to prove the claim for. This itself will be continuous. – Behnam Esmayli Apr 5 '20 at 23:27
• It seems like this method suffers from the some problem (unless I am misunderstanding). Defining such a function $f^*$ with respect to the first $M$ functions in the sequence, we get a similar inequality coming from the continuity. $0 \leq f^*(x) < f^*(x_k) + \epsilon / 2$, but then we need to find an $n_k$ such that $f_{n_k}(x_k) < \epsilon / 2$. We must also ensure that $n_k < M$. If it is not, we can increase $M$, but this changes corresponding $\delta$ and $N$. – jonan Apr 6 '20 at 1:32

Yes, it's a major trouble. Our best friend in this case is compactness

Fix $$\epsilon > 0$$.

$$\forall x \in [0, 1] \ \ \exists N_x \ \ \forall k \geq N_x : 0 < f_k(x) < \epsilon/3$$ because of pointwise convergence.

For each $$k \geq N_x$$ it exists $$\delta_{x}^{k}$$ that for each $$y \in B(x, \delta_{x}^{k}) : f_k(y) < \epsilon$$ because of continuity of $$f_k$$ (using $$|f(y)| \leq |f(x)| + |f(y) - f(x)| < \epsilon/3 \ + \epsilon/3< \epsilon)$$.

Now we consider $$\bigcup\limits_{x} \bigcup\limits_{k>N_x}B(x, \delta_{x}^{k}) \supset [0,1]$$. We extract a finite subcover using compactness.

$$\bigcup\limits_{i=1}^{j}B(x_i, \delta_{x_i}^{n_i}) \supset [0,1].$$

Now we consider $$\delta_{min} = \min\limits_{i\in{1, \ldots,j}}\delta_{x_i}^{n_i}.$$

We can easily show that each open ball $$B(x_i, \delta_{x_i}^{n_i})$$ is equal to $$\bigcup\limits_{l=1}^{p_i} B(z_l, \delta_{min})$$ for some $$z_1, \ldots, z_{p_i} \in B(x_i, \delta_{x_i}^{n_i})$$.

It appears that $$\bigcup\limits_{i=1}^{j}\bigcup\limits_{l=0}^{p_i} B(z_l, \delta_{min}) \supset [0,1]$$ QED.

• in your fourth line, do you mean $f_k$ ? – jonan Mar 30 '20 at 23:24
• Yes, sorry and it's not the same $k$ in the last lines. Now it should be OK. – Théodor Lemerle Mar 30 '20 at 23:25
• compactness? maybe you've been studying capacity on the side? – mathworker21 Mar 30 '20 at 23:29
• Of course lmao, sorry for that – Théodor Lemerle Mar 30 '20 at 23:30
• Thanks! This is very helpful – jonan Mar 31 '20 at 0:01