Integral and uniform convergence

Question

Consider $f_n(x)=nx^n$ on $[0,1)$, this is convergent pointwise to the $0$ function.

However the integral $\int_0^1 f_n =1$. If the integral is $\ne 0$ , does it imply that the function is not uniform convergent? For some reason I think it's incorrect.

How else can I prove non-uniformity?

• If $f_n$ is continuous, but $f$ that it's convergent to, fails to be continuous, it's not uniform.

• I can look for the $\sup|f_n-f|=0$, for it to be uniform.

Any other simple methods?

Answer 1

If a sequence of functions converge uniformly, then the limit and the integral commute or

$\lim_{n\rightarrow \infty}\int_{0}^{1} f_{n}d=\int_{0}^{1}\lim_{n\rightarrow \infty}f_{n}d=\int_{0}^{1}fd$

Since this is not true here, by the contrapositive of this statement, the sequence is not uniformly convergent on this interval.

Answer 2

Uniform convergence requires $\sup_{x \in [0,1)} |nx^n| \to 0$.

With $x_n = 1- 1/n \in [0,1)$ we have

$$\sup_{x \in [0,1)} |nx^n| \geqslant n(1-1/n)^n \to \infty.$$

Hence, convergence is not uniform.