# supremum of uniform bounded equicontinuous functions is continuous

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 $$\mathcal{F}$$!

• 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 $f_j(x)=1$. – Math1000 Aug 31 '19 at 21:03
• 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? – Nothingone Aug 31 '19 at 21:06

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 $$f(x) . Take supremum over $$f$$ to get $$g(x)0$$. Interchange $$x$$ and $$y$$ to get $$g(y)0$$. We have proved that $$|g(x)-g(y)| <\epsilon >0$$ whenever $$|x-y| <\delta$$.
• Cool! Just want to clarify, are we applying uniformly bounded of $\mathcal{F}$ when taking the supremum? – Nothingone Sep 1 '19 at 4:08
• @Nothingone Boundedness is required only to make sure that $g(x)<\infty$. – Kavi Rama Murthy Sep 1 '19 at 4:44