Function continuity and uniform continuity

Question

I have been through a proof that $F(x)=x^2$ is not uniformly continuous on $(-\infty,+\infty)$. Does it mean it is not continuous on $(-\infty,+\infty)$? Thanks for reading.

Answer 1

Uniform continuity is a much stronger condition than continuity.

If $f$ is uniformly continuous on $A\subseteq \Bbb R\implies$ $f$ is continuous therein.

Whereas if $f$ is not uniformly continuous on $A\subseteq \Bbb R$, nothing can be said about the continuity of $f$ on $A\subseteq \Bbb R\;$ i.e. $f$ may or may not be continuous on $A$.

Also, your function $F(x)=x^2$, is continuous on the whole of $\Bbb R$. This is how it follows.

Let $\epsilon >0$ be given and $x_0\in \Bbb R$ be arbitrary.

Consider $$ |F(x)-F(x_0)|=|x^2 -x_{0}^{2}| =|(x +x_{0})(x -x_{0})| = |x +x_{0}||x -x_{0}| \le(|x|+|x_0|)|x-x_0|$$

We now make use of the following inequality which is a consequence of the Triangle Inequality $$||a|-|b||\le|a-b|$$ For $|x-x_0|<1$, we have $$||x|-|x_0||\le|x-x_0|<1\implies|x|-|x_0|<1\implies|x|<1+|x_0|$$

and hence $$|x^2 -x_{0}^{2} | \le(|x|+|x_0|)|x-x_0|<(1+2|x_0|)|x-x_0|\tag1$$

Now choose $\delta=\min\left(1,\frac{\displaystyle\epsilon}{\displaystyle1+2|x_0|}\right)$.

Since $|x-x_0|<\delta\le 1 $ then equation $(1)$ holds, $$|x^2-x_{0}^{2}| <(1+2|x_0|)|x-x_0|$$

Again, as $\delta \le \frac{\displaystyle\epsilon}{\displaystyle(1+2|x_0|)}$

So for $|x-x_0|<\delta$, $$|F(x)-F(x_0)|=|x^2 -x_{0}^{2}| < (1+2|x_0|)|x-_0|<(1+2|x_0|)\frac{\displaystyle\epsilon}{\displaystyle(1+2|x_0|)}=\epsilon$$

Thus, $$\lim_{x\to x_0}F(x)=F(x_0)$$

Hence $F$ is continuous on $\Bbb R$.

So in this particular example we observe that $F$ is not uniformly continuous on $\Bbb R$ but it is continuous on $\Bbb R$