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If $K\subset \mathbb{R}^p$ is compact and $(f_n)$ is a sequence of continuous functions on $K$ to $\mathbb{R}^q$ which is uniformly convergent on $K$, show that the family $\{f_n\}$ is bounded on $K$.

Isn't there a theorem that states this? But, I know I must prove that there exists $M>0$ such that $||f_n(x)||\le M$ for all $x\in K$, $n\in \mathbb{N}$ which means I have to construct a sequence that converges? How will I be able to prove this?

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up vote 1 down vote accepted

$\varepsilon=1$, there being $N$, if $n\geqslant N$, then \[|f_n(x)-f_N(x)|<1 \ \forall x\in K. \] so \[|f_n(x)|<1+|f_N(x)|=M+1\ n\geqslant N,\] here, $M$ is maximun of $f_N$ on K, because $f_N$ is continuous and K is compact. And, the $N-1$ functions ahead are all bounded. We can have common maximun value, when taking the largest one of them. so sequence $f_n$ can be bounded.

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I cant read your answer, I think you forgot to use $$ to close your latex. – Q.matin Apr 20 '13 at 5:44
The answer seems to be OK. – Alex Ravsky Apr 20 '13 at 5:52
$\epsilon=1$,there being N, if $n\geq N$, then [|f_n(x)-f_N(x)|<1 \ \forall x\in K. ] so [|f_n(x)|<1+|f_N(x)|=M+1\ n\geqslant N,] here,$M $is maximun of $f_N$ on K,because $f_N$ is continuous and K is compact. And ,the $N-1$ functions ahead are all bounded.we can have common maximun value,when taking the largest one of them. so sequence$ {f_n} $can be bounded. – user63788 Apr 20 '13 at 6:04
my browser must be out of whack then because I cant read it. – Q.matin Apr 20 '13 at 6:15
use \\[ ... \\] or $$ ... $$ for displayed math. @Q.matin: it should be readable now. – Martin Apr 20 '13 at 7:21

Hint: use the limit of the sequence $\{f_n\}$ to bound it.

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Can you provide a more stronger hint? – Q.matin Apr 20 '13 at 5:26
Hint 2:Let $f$ be the limit of the sequence. Then $||f_n-f||<1$ for almost all $n$. – Alex Ravsky Apr 20 '13 at 5:28
What I got is: Let $f_n$ be a sequence that uniformly converges, then pick any integer $N$ such that $m,n\ge \mathbb{N}$ then $|f_n(x)-f_m(x)|<\epsilon$. I dont know why you chose $1$? But since $f$ is convergent continuous function it has a supremum ... – Q.matin Apr 20 '13 at 5:38

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