Checking Uniform Contuinity

Question

Check if the function is uniformly continuous: $\sqrt(x)\sin(x)$, In 0 to $\inf$ I dont think anything will happen using the definition of uniform contuinity. I cannot think of any sequence for sequential characterisation.

Answer 1

This clearly won't be uniformly continuous.

To see this, pick $\varepsilon > 0$ and try to find a $\delta > 0$, such that no matter from where in the real line you pick any $x,y \in \mathbb{R}^+$, if your $x,y$ are close enough to each other (i.e. $|x-y| < \delta$), then the distance $|f(x) - f(y)| < \varepsilon$; (where $f$ is your function).

This clearly can't happen as $\sqrt{x}$ is unbounded over the positive reals.

Suppose you have an $\varepsilon, \delta > 0$ pair you want to test. Now you pick $x_1, y_1, x_2, y_2$ such that $|x_1 - y_1|, |x_2 - y_2| < \delta$, but $x_2$ and $y_2$ are greater than $x_1, y_1$, respectively. Then the amplitude of your function $f$ will be far greater around the second set of points than the first. So the fact that $f(x_1), f(y_1)$ were $\varepsilon$-close, does not guarantee that $f(x_2), f(y_2)$ will be as well.

The key point here is that uniform continuity depends only on the value of $\varepsilon$ and not on where in the domain your points of interest are.