Uniform Convergence & Boundedness

[kingwinner]

"Let f_k be functions defined on Rⁿ 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 why it is necessarily true.

Any help is appreciated!

 

[mathman]

|f|=|f-f_k+f_k| ≤ |f-f_k| + |f_k|≤ ε + A_k

You need uniform for ε to be usable for all x.

 

[kingwinner]

Why does it imply f is bounded? Is ε fixed here?

Also, why do we need uniform convergence? (why is pointwise convergence not enough...I still don't see why)

Can someone explain this, please? Thanks!

 

[mathman]

Since the norm is defined to be the sup over all values of x
We can write:
||f||=||f-fk+fk|| ≤ ||f-fk|| + ||fk||≤ ε + Ak
Therefore ||f||≤ ε + Ak, which is a bound. ε will depend on k, but not on x.

The uniform convergence means that we can use one ε to be a bound for ||f-fk||. If it was simply pointwise convergence, ε would be a function of x and might not be bounded.

 

[kingwinner]

Sorry, I don't think it makes sense. ε is supposed to be given, always. And we're supposed to find N such that for k>N, ... will be < ε

Also, A_k depends on f_k. What we want is a bound M which is a constant and does not depend on k, right?

 

[Landau]

You need to show that f is bounded, i.e. find some positive real number M such that for all x we have |f(x)|<M. Here M cannot depend on x, but it may ver well depend on k (why not?).

mathman found such M in his first post.

 

[evagelos]

Sorry, I don't think it makes sense. ε is supposed to be given, always. And we're supposed to find N such that for k>N, ... will be < ε

Also, A_k depends on f_k. What we want is a bound M which is a constant and does not depend on k, right?

put ε =1 and k = N+1

 

[mathman]

(Uniform convergence) For every ε > 0 there exists an N so that for all k > N, ||f-fk|| < ε.
(bounded) ||fk|| < Ak. (Use any k > N)

Therefore ||f|| < ε + Ak. Thus, in plain English, f is bounded! This proof does not determine the value of the minimum bound.