Problem. Let $\mathcal{M}$ an equicontinuous and closed set of $C([a,b],\mathbb{R})$. Suppose that there is a $m \geq 0$ such that $|f(a)| \leq m$ $\forall f \in \mathcal{M}$. Show that there are $f_{1}$ and $f_{2} \in \mathcal{M}$ such that $$\int_{a}^{b}f_{1}(x)\mathrm{d}x \leq \int_{a}^{b}f(x)\mathrm{d}x \leq \int_{a}^{b}f_{2}(x)\mathrm{d}x,\quad \forall f \in \mathcal{M}.$$
My ideia is define a function $\varphi: \mathcal{M} \to \mathbb{R}$ by $\displaystyle \varphi(f) = \int_{a}^{b}f(x)dx$ and to conclude using de compactness of $\mathcal{M}$. For show that $\mathcal{M}$ is compact, I tried to use the Arzelà-Ascoli Theorem. I just to show that $\mathcal{M}(x) = \{f(x) \mid f \in \mathcal{M} \}$ is bounded and this is my problem. By hypothesis, $\mathcal{M}(a)$ is bounded and probably, I should to use it, but I don't see how. Can someone help me?
Your idea is fine. Since $\mathcal M$ is equicontinuous, there is a $\delta>0$ such that, for each $x,y\in[a,b]$ and each $f\in\mathcal M$, $|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<1$. Now, take $N\in\mathbb N$ such that $a+N\delta\geqslant b$. Then, for each $x\in[a,b]$ and each $f\in\mathcal M$, $f(x)\leqslant m+N$, because $a\leqslant x\leqslant b$ and therefore $x=a+k\delta+x'$ for some $k<N$ and some $x'\in[0,\delta]$, which implies that\begin{align}\bigl|f(x)\bigr|&=\bigl|f(a+k\delta+x')\bigr|\\&\leqslant\bigl|f(a)\bigr|+\bigl|f(a+\delta)-f(a)\bigr|+\cdots+\bigl|f(a+k\delta+x')-f(a+k\delta)\bigr|\\&\leqslant\bigl|f(a)\bigr|+k+1\\&\leqslant\bigl|f(a)\bigr|+N.\end{align}So, the hypothesis of the Arzelà-Ascoli hold and therefore $\mathcal M$ is indeed compact.