# Uniform Convergence & Boundedness

• Apr 13th 2010, 03:20 PM
kingwinner
Uniform Convergence & Boundedness
"Let $\displaystyle f_k$ be functions defined on $\displaystyle R^n$ converging uniformly to a function f. IF each $\displaystyle f_k$ is bounded, say by $\displaystyle A_k$, THEN f is bounded."

$\displaystyle f_k$ converges to f uniformly =>||$\displaystyle f_k - f$||∞ ->0 as k->∞
Also, we know|$\displaystyle f_k(x)$|≤ $\displaystyle A_k$ for all k, for all x.
But why does this imply that f is bounded? I don't see how to prove it.
Also, why do we need uniform convergence? (why is pointwise convergence not enough?)

Any help is appreciated!

[also under discussion in math links forum]
• Apr 13th 2010, 03:25 PM
Drexel28
Quote:

Originally Posted by kingwinner
"Let $\displaystyle f_k$ be functions defined on $\displaystyle R^n$ converging uniformly to a function f. IF each $\displaystyle f_k$ is bounded, say by $\displaystyle A_k$, THEN f is bounded."

$\displaystyle f_k$ converges to f uniformly =>||$\displaystyle f_k - f$||∞ ->0 as k->∞
Also, we know|$\displaystyle f_k(x)$|≤ $\displaystyle A_k$ for all k, for all x.
But why does this imply that f is bounded? I don't see how to prove it.
Also, why do we need uniform convergence? (why is pointwise convergence not enough?)

Any help is appreciated!

[also under discussion in math links forum]

Hint:
Spoiler:
$\displaystyle \|f\|_{\infty}-\|f_n\|_{\infty}\leqslant \|f-f_n\|_{\infty}$
• Apr 13th 2010, 03:43 PM
kingwinner
Quote:

Originally Posted by Drexel28
Hint:
Spoiler:
$\displaystyle \|f\|_{\infty}-\|f_n\|_{\infty}\leqslant \|f-f_n\|_{\infty}$

OK, so if we choose n large enough, we will have ||f||∞ ≤ ε + $\displaystyle A_k$. To show f is bounded, I think we can just fix some ε.

Why is pointwise convergence not enough?
• Apr 13th 2010, 03:44 PM
Drexel28
Quote:

Originally Posted by kingwinner
OK, so if we choose n large enough, we will have ||f||∞ ≤ ε + $\displaystyle A_k$. To show f is bounded, I think we can just fix some ε.

Why is pointwise convergence not enough?

Well, you need to be careful. $\displaystyle A_k$ depends on $\displaystyle f_k$, right?
• Apr 13th 2010, 04:44 PM
kingwinner
Quote:

Originally Posted by Drexel28
Well, you need to be careful. $\displaystyle A_k$ depends on $\displaystyle f_k$, right?

Why does it matter? What should we do instead? Can we take the maximum over all the A_k and use that as a bound?