# Thread: Show uniform convergence implies equicontinuity and uniform boundedness

1. ## Show uniform convergence implies equicontinuity and uniform boundedness

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.

2. Originally Posted by steph236
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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.

You can prove this using the following theorem:

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

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

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

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

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### Equicontinuous implies uniform continuity

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