1
$\begingroup$

i want to show my proof of this exercise and please if someone want review and fix my proof I will be very grateful.

Theorem: Prove that any uniformly convergent sequence of bounded functions is uniformly bounded

Proof: Let $\epsilon>0$, $\left\{ f_{n}\right\}$ a sequence defined in $A\subset R$ and $f:A\rightarrow\mathbb{R}$

As $\left\{ f_{n}\right\}$ is uniformly convergent then exist $N\in\mathbb{N}$ such that if $n\geq N$ then $|f_{n}(x)-f(x)|<1$ for all $x\in A$ (*)

Otherwise,

For all $n\in\mathbb{N}$ exists $M_{n}\in\mathbb{R}$ such that $|f_{n}(x)|<M_{n}$ for all $x\in A$

Then, if $n\geq N$ and for (*) we have:

$|f(x)|=|f_{n}(x)-f_{n}(x)+f(x)|\leq|f_{n}(x)-f(x)|+|f_{n}(x)|<1+M_{n}$

Let $C=\max\{M_{1},...,M_{n},M_{n}+1\}$ then

$|f(x)|<C$

In consequence,

$f(x)$ is uniformly bounded.

$\endgroup$
1
  • $\begingroup$ It looks like you should prove that there is a constant $C$ satisfying$|f_n(x)| \le C$ for all $x \in A$ and all $n \in \mathbb N$. $\endgroup$
    – Umberto P.
    Apr 13, 2017 at 16:45

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.