Fall 1991 #5

This begins my study of sequences and series of functions. These bad boys have a tendency of tripping me up so I’ve decided to spend some quality time learning their ways.

Problem Statement: Let A\subseteq \mathbb{R} and let \{f_n\} be a sequence of real-valued functions on A converging uniformly to a bounded real-valued function f on A. Show there exist constants K, N such that |f_n(x)|\leq K for every x\in A when n\geq N.

Proof: Let \{f_n\} be such a sequence converging uniformly to f on A where f is bounded. Since f is bounded there exists a K>0 such that |f(x)|\leq K for every x\in A.

Let \varepsilon >0, then there exists an N\in \mathbb{N} such that |f_n(x)-f(x)|<\varepsilon for every n\geq N, x\in A since \{f_n\} converges to f uniformly. Consider |f_n(x)| for n\geq N, then it follows that |f_n(x)|=|f_n(x)-f(x)+f(x)|\leq |f_n(x)-f(x)|+|f(x)|<\varepsilon+K for every x\in A. Since \varepsilon can be made arbitrarily small it follows that |f_n(x)|\leq K for every x\in A and n\geq N.


Reflection: This one wasn’t too bad. The key rests in the fact that \{f_n\}\rightarrow f uniformly, so we may say that for every \varepsilon>0 there exists an N\in \mathbb{N} such that |f_n(x)-f(x)|<\varepsilon for n\geq N and for every, x\in A. Since we may say this about every x in A we are able to make the bounding argument. If we did not have uniform convergence then we would not be able to make that argument since we would need a different N for each x\in A.

This entry was posted in Analysis, Math, Sequence of Functions, Uniform Continuity. Bookmark the permalink.

2 Responses to Fall 1991 #5

  1. j2kun says:

    You get to pick K, so it doesn’t need to be the bound on f. It could be larger, like M+11 where M is the bound on f. Then you can say that for \varepsilon = 1, there is an N with |f_n(x)| \le M + 1 < M+11 = K when n \ge N.

    I just get nervous about epsilons.

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