# Prove that the limit of a uniformly convergent sequence of bounded functions is bounded.

Prove that the limit of a uniformly convergent sequence of bounded functions from $$(E, d)$$ to $$(E', d')$$ is bounded.

$${f_n(x)}$$ is a uniformly convergent sequence of bounded functions. $$f_n \to f$$ converges uniformly.

Since $${f_n(x)}$$ is uniformly convergent, it is Cauchy. So, $$\forall \epsilon >0, \exists N \in \mathbb{N}$$ such that $$\forall n, m > N, d(f_n(x), f_m(x)) < \epsilon$$.

So, we can find an $$n$$ and $$m$$ such that $$\sup_{x \in E}|f_n(x) -f_m(x)| <1$$

I do not know how to proceed in the proof.

• Are you sure you don’t mean that a uniformly convergent sequence of bounded functions is uniformly bounded? Nov 19 '19 at 16:58
• No... the proposition I am to prove reads "Prove that the limit of a uniformly convergent sequence of bounded functions (from one metric space to another) is bounded" Nov 19 '19 at 17:04
• Your initial question never said they were bounded functions. And bounded is a weaker condition than uniformly bounded. Think of bounds for each function and then the maximum of those. Nov 19 '19 at 19:23

## 1 Answer

By uniform convergence there exists $$N \in \mathbb{N}$$ such that $$d'(f(x),f_N(x)) \leqslant 1$$ for all $$x \in E$$.

Since $$f_N$$ is bounded there exists $$y_0 \in E'$$ and finite $$r_N > 0$$ such that $$f_n(x) \in B(y_0;r_N)$$ for all $$x \in E$$.

Thus, for all $$x \in E$$ we have $$f(x) \in B(y_0,1+r_N)$$ since, by the triangle inequality,

$$d'(f(x),y_0) \leqslant d'(f(x),f_N(x)) + d'(f_N(x),y_0) \leqslant 1 + r_N$$