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Continuous function in the interval $(0,\infty)$ $f(x)=\frac{\sin(x^3)}{x}$. To prove that the function is uniformly continuous. The function is clearly continuous. Now $|f(x)-f(y)|=|\frac{\sin(x^3)}{x}-\frac{\sin(y^3)}{y}|\leq |\frac{1}{x}|+|\frac{1}{y}|$. But I don't think whether this will work.

I was trying in the other way, using Lagrange Mean Value theorem so that we can apply any Lipschitz condition or not!! but $f'(x)=3x^2\frac{\cos(x^3)}x-\frac{\sin(x^3)}{x^2}$

Any hint...

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Hint:

Any bounded, continuous function $f:(0,\infty) \to \mathbb{R}$ where $f(x) \to 0$ as $x \to 0,\infty$ is uniformly continuous. The derivative if it exists does not have to be bounded.

Note that $\sin(x^3)/x = x^2 \sin(x^3)/x^3 \to 0\cdot 1 = 0$ as $x \to 0$.

This is also a great example of a uniformly continuous function with an unbounded derivative.

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    $\begingroup$ I think in your proposition you also need $f(x)\to 0$ as $x\to 0$. Because the idea of the proof is to use that in a compact set $[\epsilon, N]$ your function is uniformly continuous and in the intervals $(0,\epsilon)$ and $(N,\infty)$ you can control the difference $|f(x)-f(y)| $since $f$ is small here. $\endgroup$ – Julian Mejia May 13 '19 at 5:27
  • $\begingroup$ @JulianMejia: That's true. Thanks. $\endgroup$ – RRL May 13 '19 at 5:31
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    $\begingroup$ Basically finite limits at both ends is what is needed, not necessarily $0$. Having overlooked that point in writing a hint gives the downvoter an opportunity. $\endgroup$ – RRL May 13 '19 at 5:34

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