Uniform Convergence & Boundedness

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SUMMARY

The discussion centers on the implications of uniform convergence for bounded functions. It establishes that if a sequence of functions \( f_k \) converges uniformly to a function \( f \) and each \( f_k \) is bounded by \( A_k \), then \( f \) is also bounded. The key argument hinges on the relationship \( ||f|| \leq \epsilon + A_k \), where \( \epsilon \) can be made arbitrarily small, thus ensuring that \( f \) remains bounded by a constant \( M \) independent of \( k \). The necessity of uniform convergence over pointwise convergence is emphasized, as uniform convergence allows for a single \( \epsilon \) applicable to all \( x \).

PREREQUISITES
  • Understanding of uniform convergence in functional analysis
  • Familiarity with bounded functions and their properties
  • Knowledge of supremum norms and their applications
  • Basic concepts of sequences and limits in mathematical analysis
NEXT STEPS
  • Study the properties of uniform convergence in more depth
  • Explore the implications of pointwise versus uniform convergence
  • Learn about supremum norms and their role in function analysis
  • Investigate examples of bounded functions and their convergence behaviors
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Mathematicians, students of analysis, and anyone studying convergence properties in functional spaces will benefit from this discussion.

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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|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!
 
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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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.
 
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
 
(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.
 

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