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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)

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    $\begingroup$ 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 $[0,∞)$. $\endgroup$
    – csch2
    Commented Mar 27, 2020 at 6:02
  • $\begingroup$ @csch2 What about my approach to (c) part? $\endgroup$ Commented Mar 27, 2020 at 6:09
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    $\begingroup$ 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. $\endgroup$
    – csch2
    Commented Mar 27, 2020 at 6:12

1 Answer 1

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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.

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  • $\begingroup$ Thanks for the help!!. I will go through 'M-test', haven't come across it yet. $\endgroup$ Commented Mar 27, 2020 at 6:15
  • $\begingroup$ 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 $\endgroup$ Commented Mar 27, 2020 at 6:27
  • $\begingroup$ @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 $x \to 0$. $\endgroup$ Commented Mar 27, 2020 at 6:30
  • $\begingroup$ @s1mple Sorry, I meant $f(x)$. $\endgroup$ Commented Mar 27, 2020 at 6:34
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    $\begingroup$ @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 $0$. $\endgroup$ Commented Mar 27, 2020 at 6:41

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