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)
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.