Show that $f(x) = x^2$ is uniformly continous on bounded set [-10, 10]

Question

I understand that if $f(x)=x^2$ is considered over any interval in general, it follows that it is uniformly continuous. Here's what I've done so far:

Given $\epsilon>0$ arbitrary, we must find a single $\delta>0$ where by $|x-y|<\delta$ $=>$ $|f(x)-f(y)|<\epsilon$

Now $|f(x)-f(y)| = |x^2-y^2|=|x+y||x-y|$

Now, we must find an upper bound on $|x+y|$ and must choose an appropriate $\delta$ such that all values of $x$ considered will fall in the interval $[-10, 10]$.

I believe I may have to use some sort of estimation on $|x+y|$, but I'm not sure what that would look like. I'm new to the concept of uniform convergence, and would greatly appreciate if someone could shed some light on how I might find a respectable $\delta$ in a situation such like the one I've outlined here. Thanks!

Answer 1

Since $\lvert x+y\rvert\leqslant20$ on $[-10,10]$, you can take $\delta=\frac\varepsilon{20}$.

Answer 2

Here's my shot at the whole proof:

Given $\epsilon>0$ arbitrary, we must find a single $\delta>0$ where by $|x-y|<\delta$ $=>$ $|f(x)-f(y)|<\epsilon$. Choose $\delta = \epsilon/20$

Now $|f(x)-f(y)| = |x^2-y^2|=|x+y||x-y| \leq 20|x-y| < 20\delta = 20(\epsilon/20)=\epsilon$