Proving $f(x)=x^2$ is uniformly continuous on $[0,b]$ for any $b\gt 0$.

Question

Consider the real function $f$ defined on the real line $\mathbb{R}$ by $f(x)=x^2$. If $b$ is a given positive real number, show that the restriction of $f$ to the closed interval $[0,b]$ is uniformly continuous by starting with an $e\gt 0$ and exhibiting a $d\gt 0$ which satisfies the requirement of the definition

Answer 1

A better way to prove this fact comes in the form of the following theorem.

Theorem. Let $K \subseteq \mathbb{R}^N$ compact, and $f:K \to \mathbb{R}$ continuous. Then $f$ is uniformly continuous.

Proof. Let $\epsilon > 0$ be given. By continuity of $f$, for every $x_0 \in K$ choose $\delta(x_0)$ such that if $x \in K$ with $\Vert x-x_0 \Vert < \delta(x_0)$ then $\Vert f(x)- f(x_0) \Vert < \epsilon/2$. The family of open balls $\{B(x,\frac{1}{2}\delta(x)) : x \in K \}$ forms an open cover of $K$. By compactness of $K$ we may choose finitely many $x_1,\ldots, x_n \in K$ such that $\{B(x_i,\frac{1}{2}\delta(x_i)) : 1 \leq i \leq n \}$ still covers $K$. Choose $\delta = \frac{1}{2}\min\{\delta(x_i) : 1 \leq i \leq n \}$. Then for any $x,y \in K$ with $\Vert x - y \Vert< \delta$ there is $i$ with $1\leq i \leq n$ so that $x \in B(x_0,\frac{1}{2}\delta(x_0))$ and $\Vert x_i - y \Vert \leq \Vert x_i - x \Vert + \Vert x-y \Vert < \frac{1}{2}\delta(x_i) + \frac{1}{2}\delta(x_i) \leq \delta(x_i)$. Thus we have $x,y \in B(x_i,\delta(x_i))$ and by the triangle inequality we have $\Vert f(x) -f(y) \Vert \leq \Vert f(x)-f(x_i) \Vert + \Vert f(x_i) - f(y) \Vert < \epsilon/2 + \epsilon/2 = \epsilon. ~~\square$

In the case of this problem, we know that $[0,b]$ is compact for all $b>0$ since it is closed and bounded, and that $x \longmapsto x^2$ is continuous on $[0,b]$. Hence by the previous theorem $x \longmapsto x^2$ is uniformly continuous on $[0,b]$.

Note: if this is a homework problem, this is probably not the solution your professor is looking for, but it is an extremely useful theorem that you might want to know about anyway.

Answer 2

If a continuous function f satisfies the property $|f(x_1)-f(x_2)|\le M|x_1-x_2|$ is said to be Lipschitz with constant $M$, and $f$ must be uniformly continuous as in this case you will get $\delta=\frac{\epsilon}{M}$ all the time which will allow your continuous function to be uniformly continuous. For your function we have $|f(x_1)-f(x_2)|=|(x_1^2-x_2^2)|=|(x_1-x_2)(x_1+x_2)|\le 2b|x_1-x_2|$. I hope you can do now.

Answer 3

Hint: for $|h|\le1$ and $x\in[0,b]$, we have $$|x^2-(x+h)^2|=|-2xh-h^2|=|h| |h+2x|\le |h|\cdot (2b+1).$$

Given $\epsilon>0$, you essentially find the "$d$ that works" for $x=b$. Then this $d$ will work for all $x\in[0,b]$.