Proving Uniform Continuity on $f(x) = 3x^2 + 2x$ on $[0, 1]$

Question

Let $\varepsilon\in\mathbb{R}^+$, and let $x,y \in [0, 1]$. Choose $\delta = \dfrac\varepsilon2$, such that $|y-x| < \delta = \dfrac\varepsilon2$. So, $|f(x) - f(y)| = \left|3y^2 + 2y - 3x^2 - 2x\right| = \left|3(y^2 - x^2) - 2(y-x\right)|$. We note that $|y-x| \le 1$ and $|y+x| \le 2$. So, $\left|3(y^2 - x^2) - 2(y-x)\right| < \left|2(y-x)\right|$. Thus, $|f(x) - f(y)| < 2|y-x| < \varepsilon$. Therefore, $|f(x) - f(y)| < \varepsilon$ so $f(x)$ is UC on $[0,1]$.

Is this proof correct?

Answer 1

There is a cleaner way to do the inequality above. Consider \begin{align*} |3(y^2-x^2)-2(y-x)|&=|3(y+x)-2||y-x|\\ &\leq |3(y+x)-2|\delta\leq (3|y+x|+2)\delta\leq8\delta \end{align*} Choosing $\delta=\epsilon/8$ will then suffice.

Answer 2

Proving uniform continuity becomes much easier(on compact sets) if you take the derivative, and apply the mean value theorem.

Essentially, just show that $f'(x) \leq n$, so $\frac{f(b)-f(a)}{b-a}<n$ for all $b,a \in S$ where the desired set is $S$. But then $f(b) - f(a) \leq \delta n$, so we just set $\delta$ to be $\frac{\epsilon}{n}$. This proves the uniform - continuity.

Here, the maximum of the derivative is $5$, so just set $\delta = \frac{\epsilon}{5}$.