# Uniform converge, using the Supremum theorem (want to make sure I solved correctly)

I have the givens: $$[a,b], [c,d] \subseteq \mathbb{R}\quad f_n:[a,b]\longrightarrow [c,d] \quad g:[c,d] \longrightarrow \mathbb{R}$$ $$g$$ is continuous and $$f_n$$ is uniformly converge on $$[a,b]$$ to $$f$$. I need to prove that $$h_n=g(f_n(x)) is$$ is uniformly converge to $$h=g(f(x))$$ on $$[a,b]$$

I showed that because $$g$$ continuous so according to heine deffinition $$\lim_{n \to \infty} g(f_n(x))=g(f(x))$$. from here I wrote $$\lim_{n \to \infty}\sup|g(f_n(x))-g(f(x))|=0.$$

The thing is I am not sure if I can argue the last equation, because I "bring" $$n$$ to infinity after I found the supremum.

Just because, for each individual $$x$$, you have $$\lim_{n\to\infty}\bigl|g\bigl(f_n(x)\bigr)-g\bigl(f(x)\bigr)\bigr|=0$$, you cannot jump to $$\lim_{n\to\infty}\sup_{x\in[a,b]}\bigl|g\bigl(f_n(x)\bigr)-g\bigl(f(x)\bigr)\bigr|=0.$$

Note that, since the domain of $$g$$ is an interval which is closed and bounded, $$g$$ is uniformly continuous. Now, take $$\varepsilon>0$$. There is a $$\delta>0$$ such that$$|x-y|<\delta\implies\bigl|g(x)-g(y)\bigr|<\varepsilon.$$And there is a $$N\in\Bbb N$$ such that$$n\geqslant N\implies(\forall x\in[a,b]):\bigl|f_n(x)-f(x)\bigr|<\delta,$$since $$(f_n)_{n\in\Bbb N}$$ converges uniformly to $$f$$. So, if $$n\geqslant N$$ and if $$x\in[a,b]$$, then$$\bigl|g\bigl(f_n(x)\bigr)-g\bigl(f(x)\bigr)\bigr|<\varepsilon.$$

• Can you please give me a numeric number why $\lim_{n\to\infty}\bigl|g\bigl(f_n(x)\bigr)-g\bigl(f(x)\bigr)\bigr|=0$ doesn't guarantee $\lim_{n\to\infty}\sup_{x\in[a,b]}\bigl|g\bigl(f_n(x)\bigr)-g\bigl(f(x)\bigr)\bigr|=0.$ ? – Sagigever May 23 at 6:46
• Take $g(x)=x$, $f_n(x)=\frac xn$ and $f(x)=0$, for instance. Then, for each $n\in\Bbb N$, $$\sup|g(f_n(x))-g(f(x)|=\sup\left|\frac xn\right|=\infty.$$ – José Carlos Santos May 23 at 6:50
• but $\lim_{n\to\infty}\sup |\frac{x}{n}|=0$ isn't? because $x \in [a,b]$ – Sagigever May 23 at 6:58
• Sure. On a bounded interval the limit is $0$ indeed. But you must justify it somehow, and you did not do that. So, the example from my previous comment shows that you must use that hypothesis (together with the continuity of all functions involved) somehow. – José Carlos Santos May 23 at 7:40

hint

Use Heine's Theorem, which states that

$$g$$ beeing continuous at the compact $$[c,d]$$, is Uniformly continuous at $$[c,d]$$.

Then, you can use Cauchy's criteria for the uniform convergence.