Find the pointwise limit and the uniform convergence of $f_n(x) = x(1-x^n)$ and $f_n(x) = \frac{nx}{n+2}$

Question

Find the pointwise limit and the uniform convergence of $f_n(x) = x(1-x^n)$ and $f_n(x) = \frac{nx}{n+2}$ where for both functions$f_n : [0,1] \to \mathbb{R} $

For the first function, I found the pointwise limit to be: $ x $ for $x \in [0, 1)$ and $0$ if $x=1$

To determine uniform convergence, $$f_n'(x) = (1-x^n) + x(-nx^{n-1}) = 1-x^n -nx^n = 1 - (1+n)x^n$$

I am having trouble moving forward determining uniform convergence.

For the second function, the pointwise limit is x for all x. I also do not how to go on and prove uniform convergence

Answer 1

A useful theorem: if $(f_n)_n$ is a sequence of continuous functions converging uniformly to a function $f$, then $f$ is continuous. That will take care of the first:

• The pointwise limit of $(f_n)_n$ is $f$
• if $(f_n)_n$ converges uniformly, then its uniform limit is the same as its pointwise limit
• for every $n$, $f_n$ is continuous on $[0,1]$
• but... $f$ here is not exactly continuous on $[0,1]$, is it?

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As for the second: for every $x\in[0,1]$, and all $n$ $$ f(x) - f_n(x) = \frac{(n+2)x}{n+2} - \frac{n x}{n+2} = \frac{2x}{n+2} $$ so that $$ \sup_{x\in[0,1] }\lvert f(x) - f_n(x)\rvert = \sup_{x\in[0,1] }\left\lvert \frac{2x}{n+2}\right\rvert = \frac{2}{n+2} \xrightarrow[n\to\infty]{} 0 $$ so you can now conclude about uniform convergence.