# Math Help - Grade My Uniform Continuity Proof

1. ## Grade My Uniform Continuity Proof

The goal is to show that if $fa,b)\rightarrow \mathbb{R}" alt="fa,b)\rightarrow \mathbb{R}" /> is uniformly continuous, then $f$ is bounded.

Here is my attempt...
Proof: Let $fa,b)\rightarrow \mathbb{R}" alt="fa,b)\rightarrow \mathbb{R}" /> be uniformly continuous. Then, by definition, given any $\epsilon >0$ there exists a $\delta >0$ such that for all $x,y\in (a,b), |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$. Thus, f is continuous on $[x,y]$. Therefore, since $[x,y]$ is closed and bounded, and $f:[x,y]\rightarrow \mathbb{C}$ is contiuous on $[x,y]$, f is bounded on [x,y].

I'll take any constructive criticism that you want to give me. Thanks.

2. Originally Posted by zebra2147
Therefore, since $[x,y]$ is closed and bounded, and $f:[x,y]\rightarrow \mathbb{C}$ is contiuous on $[x,y]$, f is bounded on [x,y].
Should be $f:[x,y]\rightarrow \mathbb{R}$.

The proof is so simple that to grade it one has to know what you are allowed to use and what level of details you have to provide. It is not clear, for example, why you can use the theorem that continuous functions are bounded but cannot use the fact that uniformly continuous functions are continuous.

3. Originally Posted by zebra2147
The goal is to show that if $fa,b)\rightarrow \mathbb{R}" alt="fa,b)\rightarrow \mathbb{R}" /> is uniformly continuous, then $f$ is bounded.

Here is my attempt...
Proof: Let $fa,b)\rightarrow \mathbb{R}" alt="fa,b)\rightarrow \mathbb{R}" /> be uniformly continuous. Then, by definition, given any $\epsilon >0$ there exists a $\delta >0$ such that for all $x,y\in (a,b), |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$. Thus, f is continuous on $[x,y]$. Therefore, since $[x,y]$ is closed and bounded, and $f:[x,y]\rightarrow \mathbb{C}$ is contiuous on $[x,y]$, f is bounded on [x,y].

I'll take any constructive criticism that you want to give me. Thanks.
Where did that closed interval come from? And why does it show anything?

There are two methods of attack that I see.

1) Use heavy machinery to say that $fa,b)\to\mathbb{R}" alt="fa,b)\to\mathbb{R}" /> may be extended to some uniformly continuous function $f:[a,b]\to\mathbb{R}$ and every continuous (real valued) function on a compact set is bounded.

2) If you can't use any big guns suppose that $f$ was unbounded. Then, there is a sequence of points $\{x_n\}_{n\in\mathbb{N}}$ such that $f(x_n)>n$. But, since $\{x_n\}_{n\in\mathbb{N}}$. But, every bounded sequence of numbers has a convergent subsequence (it's stuck inside some compact set) and so in particular $\{x_{n}\}_{n\in\mathbb{N}}$ has some convergent subsequence $\{x_{n_k}\}_{k\in\mathbb{N}}$ and in particular it's Cauchy. But, uniformly continuous functions preserve Cauchy sequences and thus $\{f(x_{n_k})\}_{k\in\mathbb{N}}$ is Cauchy. But, this is impossible since $f(x_{n_k}>n_k$.