Application of the Uniform limit theorem

Calculus Level 1

For n ∈ N n\in \mathbb{N} , let f n : [ 0 , π ] → R f_n:[0, \pi]\to \mathbb{R} be defined by

f n ( x ) : = sin ⁡ n ( x ) f_n(x):= \sin^n (x)

Is the sequence { f n } n = 1 ∞ \{f_n\}_{n=1}^{\infty} uniformly convergent?

Hint : is the limit function continuous?

This problem is often posed in calculus textbooks.
No Yes

Ognjen Vukadin by Ognjen Vukadin , 38, Austria

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1 solution

Ognjen Vukadin
Jul 15, 2016

The limit function is discontinuous:

f ( x ) = { 1 , x = π 2 0 , x ∈ [ 0 , π ] ∖ { π 2 } f(x) = \begin{cases} 1, & x = \frac{\pi}{2} \\ 0, & x\in [0,\pi]\setminus \{\frac{\pi}{2}\} \\ \end{cases}

Uniform limit theorem implies that this can happen only if convergence is not uniform since all the functions f n f_n are continuous.

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