# Thread: Uniform Convergence & Boundedness

1. ## Uniform Convergence & Boundedness

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

$f_k$ converges to f uniformly =>|| $f_k - f$||∞ ->0 as k->∞
Also, we know| $f_k(x)$|≤ $A_k$ 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]

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

$f_k$ converges to f uniformly =>|| $f_k - f$||∞ ->0 as k->∞
Also, we know| $f_k(x)$|≤ $A_k$ 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:
$\|f\|_{\infty}-\|f_n\|_{\infty}\leqslant \|f-f_n\|_{\infty}$

3. Originally Posted by Drexel28
Hint:
Spoiler:
$\|f\|_{\infty}-\|f_n\|_{\infty}\leqslant \|f-f_n\|_{\infty}$
OK, so if we choose n large enough, we will have ||f||∞ ≤ ε + $A_k$. To show f is bounded, I think we can just fix some ε.

Why is pointwise convergence not enough?

4. Originally Posted by kingwinner
OK, so if we choose n large enough, we will have ||f||∞ ≤ ε + $A_k$. 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. $A_k$ depends on $f_k$, right?

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