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$\left \{f_n \right \}$ 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.

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  • $\begingroup$ 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. $\endgroup$ – Behnam Esmayli Apr 5 '20 at 23:27
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    $\begingroup$ 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 $N$. $\endgroup$ – jonan Apr 6 '20 at 1:32
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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.

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

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