Uniform convergent sequences of functions?

Calculus Level 2

Which of the following sequences of functions $\{f_n\}_{n=1}^{\infty}$ is/are uniformly convergent on its/their domain?

(A) : $f_n(x) = \cos^{n}(x)$ on $\left[-\dfrac \pi2, \dfrac \pi2\right]$ .

(B) : $f_n(x) = \dfrac{x}{n}$ on $\mathbb{R}$ .

(C) : $f_n(x) = \dfrac{\sin(nx)}{n}$ on $\mathbb{R}$ .

Notation : $\mathbb R$ denotes the set of real numbers .

(B) only (B) and (C) only (A) only (C) only

1 solution

Ognjen Vukadin
Jul 17, 2016

Relevant wiki: Uniform Convergence

A: All the functions $f_{n}$ are continuous but the pointwise limit function $f(x): [-\pi/2, \pi/2]\to \mathbb{R}$ is discontinuous:
$f(x) = \begin{cases} 1, & x = 0 \\ 0, & otherwise \\ \end{cases}$ The Uniform limit theorem implies that the convergence can not be uniform.

B: The functions $f_n$ converge pointwise to the constant function $f(x) = 0$ on $\mathbb{R}$ . However, $\displaystyle\lim_{n\to \infty}||f_n - 0||_{\infty}\neq 0$ on $\mathbb{R}$ so the convergence is not uniform.

C: The functions $f_n$ converge pointwise to the constant function $f(x)=0$ on $\mathbb{R}$ and $\displaystyle\lim_{n\to \infty}||f_n - 0||_{\infty} = \displaystyle\lim_{n\to \infty} \frac{1}{n} = 0$ on $\mathbb{R}$ so the convergence is uniform.

Uniform convergent sequences of functions?

1 solution

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