Let $f_n(x)=nx(1-x^2)^n$ on $[0,1]$ for $n\ge1$. Find $f(x)= \lim f_n(x)$. Is this a convergence uniform?

Question

Let $f_n(x)=nx(1-x^2)^n$ on $[0,1]$ for $n\ge1$.

Find $f(x)= \lim f_n(x)$.

Is this a convergence uniform?

I have to use the hint that $\lim_{n\to \infty}(1-\frac hn)^n=e^{-h}$.

Do we show $|f(x)- f_n(x)| < \epsilon$ and go from there?

Answer 1

For $f_n(x) = nx(1 -x^2)^n$, the sequence converges pointwise to $0$ when $x \in [0,1]$ since $f_n(0) = f_n(1) = 0,$ and for $0 < x < 1$,

$$0 \leqslant f_n(x) < n\alpha^n= ne^{-|\ln \alpha|n},$$

where $\alpha < 1$.

A maximum is attained at $x = 1/\sqrt{2n+1},$ and

$$\sup_{x \in [0,1]}|f_n(x)|= \frac{n}{\sqrt{2n + 1}}\left(1-\frac{1}{2n+1}\right)^n.$$

Note that

$$\lim_{n \rightarrow \infty}\left(1-\frac{1}{2n+1}\right)^n= \lim_{n \rightarrow \infty}\left[\left(1-\frac{1}{2n+1}\right)^{2n+1}\right]^{1/2}\left(\frac{2}{2+1/n}\right)^{-1/2}=e^{1/2}.$$

For $n$ sufficiently large,

$$\left(1-\frac{1}{2n+1}\right)^n > \frac{e^{1/2}}{2},$$

and

$$\sup_{x \in [0,1]}|f_n(x)| > \frac{\sqrt{n}}{\sqrt{2 + 1/n}}\frac{e^{1/2}}{2}$$

Hence,

$$\lim_{n \rightarrow \infty}\sup_{x \in [0,1]}|f_n(x)| = \infty,$$

and the convergence is not uniform.