# Which of the following are uniformly continuous?

Which of the following functions are uniformly continuous over their respective domain of definition?

(a) $$f(x) = \text{cos}x\text{cos}\frac{\pi}{x}$$, $$x\in(0,1)$$

(b) $$g(x) = \text{sin}x\text{sin}\frac{\pi}{x}$$, $$x\in(0,1)$$

(c) $$h(x) = \sum_{n=1}^{\infty} \frac{g(x-n)}{2^n},\; x \in \mathbb{R}$$, where $$g:\mathbb{R}\to\mathbb{R}$$ is a bounded uniformly continuous function.

My attempt:

Theorem: Any function which is differentiable and has bounded derivative is uniformly continuous (this follows from the MVT).

Now, their derivatives turn out to be: $$f'(x) = -\text{sin}x\text{cos}\frac{\pi}{x}+\pi \text{sin}(\frac{\pi}{x})cos(x)\frac{1}{x^2}$$ and $$g'(x)=\text{cos}x\text{sin}\frac{\pi}{x}-\pi \text{cos}(\frac{\pi}{x})sin(x)\frac{1}{x^2}$$

Clearly $$\lim_{x\to 0}f'(x)$$ and $$\lim_{x\to 0}g'(x)$$ unbounded because of the presence of $$\frac{1}{x^2}$$ term.

Now, coming to option (c), we know that sum of two uniformly cont. functions is again uniformly continuous (in general,product and quotient are not). Also if $$g(x)$$ is uniformly continuous then so is $$g(x-n)$$ and so $$\frac{g(x-n)}{2^n}$$ being uniformly continuous, the function $$h(x)$$ is option (c) is uniformly continuous.

Is my approach to the above problems correct?

I am getting only option (c) as correct answer.

The answers are option : (b) and (c)

• Bounded derivative gives uniform continuity, but an unbounded derivative is not enough to conclude that a function is not uniformly continuous. Take $\sqrt x$ on $[0,∞)$. Commented Mar 27, 2020 at 6:02
• @csch2 What about my approach to (c) part? Commented Mar 27, 2020 at 6:09
• You need to be careful. While it is true that a finite sum of uniformly continuous functions is uniformly continuous, this may not hold for infinite sums. However, the uniform limit of a sequence of uniformly continuous functions is uniformly continuous, so if you can prove that your infinite sum converges uniformly (which is simple, if you exploit the boundedness of $g$ and use Weierstrauss M-test), then the resulting infinite sum of uniformly continuous functions is also uniformly continuous. Commented Mar 27, 2020 at 6:12

## 1 Answer

The theorem you have quoted works only one way so it is not good enough to answer this question. Here are some hints:

A continuous function on $$(0,1)$$ is uniformly continuous iff it has a finite limit at $$0$$ and $$1$$. In a) the function does not have a finite limit at $$0$$. [Look at points where $$\cos (\frac {\pi} x)$$ has the values $$0$$ and $$1$$].

In b) the limits do exist. Note that $$|\sin x \sin (\frac {\pi} x)| \leq |\sin x| \to 0$$ as $$x \to 0$$.

In c) use the fact that the series is uniformly convergent (by M-test). So the partial sums converge uniformly and each partial sum is uniformly continuous. This implies that the sum is uniformly continuous.

• Thanks for the help!!. I will go through 'M-test', haven't come across it yet. Commented Mar 27, 2020 at 6:15
• cos($\frac{\pi}{x}$) has value 1 as $x\to \infty$ and 0 when $x=\frac{2}{2n+1}$, so in (0,1) cos$(\frac{\pi}{x})$ has value 0 at infinitely many points Commented Mar 27, 2020 at 6:27
• @s1mple That is correct but you also need the fact that the function has the value $1$ at $x=\frac 1 {2n}$. This shows that $g(x)$ does not have a limit as $x \to 0$. Commented Mar 27, 2020 at 6:30
• @s1mple Sorry, I meant $f(x)$. Commented Mar 27, 2020 at 6:34
• @s1mple $\cos (\frac {\pi} x$ oscillates between $-1$ and $+1$. It does not have a limit at $0$. For a function$f(x)$ to have a limit $l$ as $x \to 0$ it is necessary that $f(x_n) \to l$ for all sequences tending to $0$. When you get different limits for different sequences tending to the function has no limit at $0$. Commented Mar 27, 2020 at 6:41