I am working on a problem in Berkeley book but it has no solution. I want someone give me comments on my proof.
Here is the problem : Let $\mathcal{F}$ be a uniformly bounded, equicontinuous family of real-valued functions on a metric space $(X,d)$. Prove that $g(x)=\sup\{f(x): f\in\mathcal{F}\}$ is continuous.
Proof : First, we notice that $g(x)=f_j(x)$ for some $f_j\in\mathcal{F}$ and $g(y)=f_k(y)$ for some $f_k\in\mathcal{F}$. Thus we have
\begin{align} g(x)-g(y)&=f_j(x)-f_k(y)\\&=f_j(x)-f_j(y)+f_j(y)-f_k(y)\\&\leq f_j(x)-f_j(y) \quad(\text{since}\,\, f_j(y)\leq f_k(y))\end{align}
Since $\mathcal{F}$ is an equicontinuous family, we can make the following inequality arbitrary small
$$|g(x)-g(y)|\leq |f_j(x)-f_j(y)|<\varepsilon$$
It turns out that I did not use the assumption that the family $\mathcal{F}$ is uniform boundedness. Did I miss some details?
Edit: It seems my proof does not work according to the below discussion. Does anyone have the right idea to tackle the problem?
My second attempt: Take $g(x)=p(x)$ and $g(y)=q(y)$ for some real-valued functions $p(x)$ and $q(x)$ on $X$. Let $\delta>0$. Then we have $p(x)-\delta\leq g(x)$. Also, we have $f(y)\leq q(y)$ which implies $-q(y)\leq -f(y).$ Thus we have
\begin{align} g(x)-g(y)&=p(x)-q(y)\\&< f(x)+\delta-f(y)\\&\leq f(x)-f(y)\quad(\text{since}\,\,\,\delta\,\,\,\text{is arbitrary}) \end{align}
Hence the result follows! Again, this approach I did not use the uniformly bounded family of $\mathcal{F}$!
Best Answer
As pointed out already your proof doesn't work.
By definition of equicontinuity, given $\epsilon >0$, there exists $\delta >0$ such that $|f(x)-f(y)| <\epsilon$ whenever $|x-y| <\delta$ and $f \in \mathcal F$. Hence $f(x) <f(y)+ \epsilon \leq g(y)+\epsilon $. Take supremum over $f$ to get $g(x)<g(y)+\epsilon >0$. Interchange $x$ and $y$ to get $g(y)<g(x)+\epsilon >0$. We have proved that $|g(x)-g(y)| <\epsilon >0$ whenever $|x-y| <\delta$.