Question

Show that f(x)=x^2 is not uniformly continuous on R.

Answer 1

Show that $f(x) = x^2$ is not uniformly continuous on $R$ . Solution.

Function $f(x)$ is uniformly continuous on $R$ if for arbitrary $\varepsilon > 0$ there exists $\delta > 0$ such that as soon as $|x - \chi_0| < \delta$ it follows that $|f(x) - f(\chi_0)| < \varepsilon$ for any $\chi_0 \in R$ .

Note that the value of $\delta$ doesn't depend upon the choice of $\chi_0$ .

Now suppose that $\varepsilon > 0$ is given and let us try to find corresponding value of $\delta$ .

Using Lagrange theorem we can write that

$$\left| f (x) - f (\chi) _ {0} \right| = \left| \tilde {f} ^ {\prime} (c) \right| ^ {\frac {1}{2}} \left| x - \chi_ {0} \right| = \left| 2 c \right| \left| x - \chi_ {0} \right|,$$

where $c\in [\chi_0,x]$

Let $\left|x - \chi_0\right| < \delta$ and in order to find the value of $\delta$ we demand that

$$\left| 2 c \right| \left| x - \chi_ {0} \right| < \left| 2 c \right| \delta = \varepsilon .$$

From this $\delta = \varepsilon / |2c|$ . But as the value of $c$ depends on $\chi_0$ then the value of $\delta$ depends on the choice of $\chi_0$ too. So the value of $\delta$ which is common for all points $\chi_0 \in R$ doesn't exist.

Therefore the function $f(x) = \chi^2$ is not uniformly continuous on $R$ . Q.E.D.