🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

We first prove that for any $x_0\in [a,b]$ and $ϵ\in {\mathbb{R}}^{+}$ there exists some integer $N\in {\mathbb{N}}_1$ and $r\in {\mathbb{R}}^{+}$ such that $${\displaystyle |x-x_0|\le r⟹\left|g_N\left(x\right)\right|\le ϵ}$$ where $g_n\left(x\right)=f\left(x\right)-f_n\left(x\right)$

This follows from a few facts, we first start with continuity of $f,f_n$, specifically at $x_0$ and therefore using $ϵ$ we obtain some ${\delta }_1\in {\mathbb{R}}^{+}$ such that $f\left(B(x_0,{\delta }_1)\right)\subseteq B(g_n\left(x_0\right),ϵ)$, similarly we have a ${\delta }_2\in {\mathbb{R}}^{+}$ such that $f\left(B(x_0,{\delta }_2)\right)\subseteq B(f\left(x_0\right),ϵ)$, then also we know that since $f_n\mathrm{\backslash stackrel}pw\to f$ then since it's true at $x_0$ and using $ϵ$ we obtain some $K\in {\mathbb{N}}_1$ such that for all $n\in {\mathbb{N}}_K$ we have $\left|f_n\left(x_0\right)\right|-f\left(x_0\right)\le ϵ$

Now we combine using the triangle inequality, so let $ϵ\in {\mathbb{R}}^{+}$ , then let $r=min({\delta }_1,{\delta }_2)$ so that we obtain the inequalities mentioned in the previous paragraph, and we get some $K\in {\mathbb{N}}_1$ as mentioned in the previous paragrah, then $$\begin{gathered} {\displaystyle \left|g_K\left(x\right)\right|\&=|f\left(x\right)-f_K\left(x\right)| \\ \&\le |f\left(x\right)-f\left(x_0\right)|+|f\left(x_0\right)-f_K\left(x_0\right)|+|f_K\left(x_0\right)-f_K\left(x\right)| \\ \&\le 3ϵ} \end{gathered}$$

With that out of the way suppose for the sake of contradiction that $f_n$ doesn't convege uniformly to $f$, since $f-f_n$ is eventually bounded then this is equivalent to $${\displaystyle {lim}_{n\to \infty }\Vert f-f_n{\Vert }_{\infty }=d\in {\mathbb{R}}^{+}}$$

Since $g_n$ is continuous as the sum of continuous functions and has compact domain then by EVT, we know that it attains its max and min, and therefore there exists a $x_k\in [a,b]$ such that $\Vert f-f_k{\Vert }_{\infty }=g_k\left(x_k\right)$, so we have ${lim}_{k\to \infty }{g}_k\left(x_k\right)=d$, and note that this means that for any ${ϵ}^{\prime }\in {\mathbb{R}}^{+}$ there is some $J\in {\mathbb{N}}_1$ such that for all $j\in {\mathbb{N}}_J$ we have $|g_j\left(x_j\right)-d|<{ϵ}^{\prime }$ which implies that $d-{ϵ}^{\prime }\le g_j\left(x_j\right)$.

Since $x_k$ is in a compact set it has a convergent subsequence $x_{n_k}\to x\in [a,b]$. Now recall the fact we proved initially, by using $d-ϵ$ we know that there is some $N\in {\mathbb{N}}_1$ such that $g_N\left(x\right)<d-ϵ$, now at the same time the inequality discussed in the previous paragraph also holds on the subsequence (this is because if a sequence goes somewhere, then any subsequence must also go there), namely we have some $J\in {\mathbb{N}}_1$ such that for all $j\in {\mathbb{N}}_J$ $${\displaystyle d-ϵ\le g_{n_j}\left(x_{n_j}\right)}$$ but then there is some $i$ such that $n_i>max(N,J)$ and therefore $g_i<g_N$ and so chaining inequalities we have $${\displaystyle d-ϵ\le g_{n_i}\left(x_{x_i}\right)\le g_N\left(x\right)\le d-ϵ}$$ which is a contradiction, and therefore $f_n\mathrm{\backslash stackrel}unif\to f$