$K$ be compact metric space, $g$ and $h$ are continuous, prove the family of $f_u(x) = h(g(x)+g(u))$ is equicontinuous.

Question

$K$ be compact metric space and $g: K \rightarrow R$ be continuous. $\forall u \in K$ let $f_u : K \rightarrow R$ be $f_u(x) = h(g(x)+g(u))$, where $h$ is any continuous function on $R$, show that the family of $f_u(x)$ is equicontinuous.

My intuition: $\forall u$ $g(u)$ is bounded and fixed. so we can replace $g(u)$ with $C$. $hg(x)$ is continuous, so $h(g() +C)$ is continuous. Thus, $ℎ(()+())$ is continuous, and $f_u$ is equicontinuous.

Is my intuition correct?

Thanks

Answer 1

No. Your intuition is not correct. Here is why:

By definition, a family $F$ of (real) functions defined on a metric space $(K,d)$ is said to be equicontinuous if for every given $\epsilon>0$ there exists one single number $\delta>0$ depending only on $\epsilon$ satisfying the following: for every given triple $(x,y,F) \in K \times K \times F$ such that $d(x,y)<\delta$, we always have $|f(x)-f(y)|<\epsilon$.

In the context of your question, this means: for a given $\epsilon>0$, you must find one single number $\delta > 0$ such that for every $(x,y,u) \in K^3$such that $d(x,y)<\delta$ we have $|f_u(x)-f_u(y)|<\epsilon$. The chosen $\delta$ must not depend on $u$. Therefore, you cannot just replace $g(u)$ as a constant as that would only prove $f_u$ is continuous for every fixed given $u \in K$. That is, in your solution, your $\delta$ depends on $u$.

To fix this, note that since the domain of $g$ is compact, $g$ is bounded and $g$ is uniformly continuous on $K$. There exists an $M>0$ such that $g(x)+g(u) \in [-M,M]$ for every $x \in K$ and every $u \in K$. In particular, $h$ is uniformly continuous on $[-M,M]$. Therefore, you can approach the problem as follows:

Let $\epsilon>0$ be given. There exists a $\delta_1>0$ such that $|h(a)-h(b)|<\epsilon$ whenever $|a-b|<\delta_1$ and $a,b \in [-M,M]$ by uniform continuity of $h$. By uniform continuity of $g$, there exists a $\delta > 0$ such that $|g(x)-g(y)|< \delta_1$ whenever $d(x,y)<\delta$. This is indeed the desired $\delta$ as $|g(x)-g(y)|=|g(x)+g(u)-g(y)-g(u)|$ so that you put $a=g(x)+g(u)$ and $b=g(y)+g(u)$ the proof is complete.