Show uniform convergence implies equicontinuity and uniform boundedness

**steph236** · October 30th 2010, 01:14 PM

Hi guys! I was wondering if you could look at my proof and tell me if you think its correct/rigorous enough. I'm having a little difficulty with this.

The question is: Let (f_n) be a sequence of functions in C([0,1]), with f_n uniformly converging to f in [0,1]. Show without using the Heine-Borel theorem that the sequence is also uniformly bounded and equicontinuous in [0,1].

Equicontinuity:
By uniform convergence, the sequence of functions meets the Caunchy criterion, so there exists a natural number N such that when n,m>=N:

|f_n(x)-f_m(x)|<E/3 (for every x in [0,1])

By continuity of each f_k, there exists a d such that:

|x-y|<d implies |f_k(x)-f_k(y)|<e/3 (for every x,y in [0,1])

Combining these two facts, we have that for any e>0, when n,k>=N and |x-y|<d:

|f_n(x)-f_n(y)| =< |f_n(x)-f_k(x)| + |f_k(x)-f_k(y)| + |f_k(y)-f_n(y)|
|f_n(x)-f_n(y)| =< e/3 + e/3 + e/3
|f_n(x)-f_n(y)| =< e (for every x,y in [0,1])

Which is exactly the definition of equicontinuity. So, f_n is equicontinuous on [0,1].

Uniform Boundedness:
Since each f_k is continuous, and [0,1] is closed and bounded, by the Extreme Value theorem, each f_k attains its max and min on [0,1], ie. they are each bounded. Let c=max{|f_1(x)|, |f_2(x)|, |f_3(x)|,...}. Then |f_n(x)|=<c.

I don't think this last part works, but I'm not sure what else to do. Maybe something like:

If it were unbounded, then in the function space C([0,1], R) for any e>0 sup|f_k(x)-f(x)|>e for some value of x. But that contradicts what uniform convergence tells us, ie. that theres an N s.t k>=N means sup|f_k(x)-f(x)|<e for all x.

Any comments/helpful tips are much appreciated. =)

**Drexel28** · October 31st 2010, 07:09 PM

Originally Posted by steph236

H
Uniform Boundedness:
Since each f_k is continuous, and [0,1] is closed and bounded, by the Extreme Value theorem, each f_k attains its max and min on [0,1], ie. they are each bounded. Let c=max{|f_1(x)|, |f_2(x)|, |f_3(x)|,...}. Then |f_n(x)|=<c.

I don't think this last part works, but I'm not sure what else to do. Maybe something like:

If it were unbounded, then in the function space C([0,1], R) for any e>0 sup|f_k(x)-f(x)|>e for some value of x. But that contradicts what uniform convergence tells us, ie. that theres an N s.t k>=N means sup|f_k(x)-f(x)|<e for all x.

Any comments/helpful tips are much appreciated. =)

You can prove this using the following theorem:

Theorem: Let $\{f_n\}_{n\in\mathbb{N}}$ be a sequence of bounded on $[a,b]$ functions and $f_n{\xrightarrow{\text{unif.}}}f$ with $f$ bouned. Then, $\{f_n\}_{n\in\mathbb{N}}$ is uniformly bounded.

Proof: We know that $f_n{\xrightarrow{\text{unif.}}f$ if and only if $f_n\to f$ with the $\|\cdot\|_{\infty}$ norm. So, choose $N\in\mathbb{N}$ such that $N\leqslant n\implies \|f_n-f\|_{\infty}<1$ . Then, let $M=\max\{\|f_1\|_{\infty},\cdots,\|f_{N-1}\|_{\infty},\|f\|_{\infty}+1\}$ . Then, let $f_k$ be arbitrary. If $k<N$ then evidently $\|f_k\|_{\infty}<M$ if $k\geqslant N$ then $\|f_k\|_{\infty}-\|f\|_{\infty}\leqslant \|f_k-f\|_{\infty}<1\implies \|f_k\|<\|f\|_{\infty}+1\leqslant M$ . Regardless, for every $n\in\mathbb{N}$ we have that $\|f_k\|_{\infty}\leqslant M$ . Or, that for every $x\in[a,b]$ and every $n\in\mathbb{N}$ we have that $|f_n(x)|\leqslant M$ . $\blacksquare$

Remark: The above proof is not necessarily the most general but will work for our purposes.

Thus, noticing that the continuity of $f_n$ and compactness of $[0,1]$ guarantees that each $f_n$ is bounded, and since the uniform limit of continuous functions is continuous we see that $f$ is a continuous function on a compact space and thus also bounded. Thus, the theorem applies

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