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