# Uniform Convergence & Boundedness

"Let fk be functions defined on Rn converging uniformly to a function f. IF each fk is bounded, say by Ak, THEN f is bounded."

fk converges to f uniformly =>||fk - f|| ->0 as k->∞
Also, we know|fk(x)|≤ Ak 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!

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mathman
|f|=|f-fk+fk| ≤ |f-fk| + |fk|≤ ε + Ak

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

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.

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

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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.

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, Ak depends on fk. 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