Uniform Convergence & Boundedness

In summary, if functions defined on Rn converge uniformly to a function f and each individual function is bounded by Ak, then f is also bounded. This is because the uniform convergence allows us to use one fixed value for ε, which in turn helps us find a constant bound for f.
  • #1
kingwinner
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"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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  • #2
|f|=|f-fk+fk| ≤ |f-fk| + |fk|≤ ε + Ak

You need uniform for ε to be usable for all x.
 
  • #3
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!
 
  • #4
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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  • #5
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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  • #6
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.
 
  • #7
kingwinner said:
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
 
  • #8
(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.
 

1. What is uniform convergence?

Uniform convergence is a type of convergence in which the difference between the values of a sequence of functions and the limit function approaches zero as the input value approaches infinity. This means that the convergence is independent of the input value and holds true for all values within a given interval.

2. How is uniform convergence different from pointwise convergence?

Pointwise convergence only requires that the limit of a sequence of functions approaches the limit function at each point individually. In contrast, uniform convergence requires that this convergence holds true for all points within a given interval simultaneously.

3. Can a uniformly convergent sequence of functions be unbounded?

Yes, a sequence of functions can be uniformly convergent but still be unbounded. This means that while the differences between the values of the functions and the limit function approach zero, the values of the functions themselves may still become infinitely large.

4. How does boundedness relate to uniform convergence?

Boundedness is a necessary condition for uniform convergence. This means that in order for a sequence of functions to be uniformly convergent, it must also be bounded. If a sequence of functions is not bounded, it cannot be uniformly convergent.

5. What are the implications of uniform convergence for the continuity of a function?

If a sequence of functions is uniformly convergent, the limit function will also be continuous. This is because uniform convergence ensures that the differences between the values of the functions and the limit function will approach zero, allowing for a smooth transition between the values of the functions and the limit function.

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