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I remember proving $x^3$ is not uniformly continuous on $\mathbb{R}$. Then I read the proof of a theorem: Suppose $D$ is compact. Function $f: D \rightarrow \mathbb{R}$ is continuous on $D$ if and only if f is uniformly continuous.

It's obvious that $f(x)=x^3$ is continuous at $x$ for all $x\in \mathbb{R}$

So does it mean if I bounded the interval $f(x)=x^3$ to, let's say, $[a,b]$ with $a<b$, then $f(x)=x^3$ is uniformly continuous?

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Yes. Exactly. Uniform continuity asserts a global bound on how close you need to look in order to approximate $f$ by a certain amount; intuitively, on an infinite set there might be no global bound because the function might run off to infinity faster and faster as time goes on, while on a closed and bounded set it's prevented from doing that.

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Since your function is a polynomial and hence derivable.Thus using the fact if the derivative exists and is bounded then it is is uniformly continuous. $f(x)=x^3$, $f'(x)=3x^2$ and it is bounded on your interval $|f'(x)|\le 3b^2$ for all $x$ in $[a,b]$

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