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 $\mathcal{F}$!

  • $\begingroup$ It is not the case that $g(x)=f_j(x)$ for some $f_j\in\mathcal F$. Take the members of $\mathcal F$ to be $f_n(x) = 1-\frac1n$. Then $g(x)=1$ but there is no $f_j\in\mathcal F$ with $f_j(x)=1$. $\endgroup$ – Math1000 Aug 31 '19 at 21:03
  • $\begingroup$ Well ,you're right. It is not necessary that the supremum will be one of those in the family $\mathcal{F}$. Do you have any idea to fix it? $\endgroup$ – Nothingone Aug 31 '19 at 21:06

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 $|x-y| <\delta$.

  • $\begingroup$ Cool! Just want to clarify, are we applying uniformly bounded of $\mathcal{F}$ when taking the supremum? $\endgroup$ – Nothingone Sep 1 '19 at 4:08
  • $\begingroup$ @Nothingone Boundedness is required only to make sure that $g(x)<\infty$. $\endgroup$ – Kavi Rama Murthy Sep 1 '19 at 4:44

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