Suppose $f_n:[0,1]\to\mathbb{R}$ such that $f_n=nx(1-x)^n$, then I need to show that $f_n$ converges pointwise but not uniformly. I can see that since $f$ is defined on $[0,1]$, the part $(1-x)^n$ is smaller than $1$ and therefore will converge, but what about $nx$? This will go to infinity…
Therefore, I can't even see that it converges pointwise, but let's suppose it converges to some function, how do I show that it doesn't converge uniformly? I should find an $\epsilon$ such that it's not valid for all $n>n_0(\epsilon)$ that $|f_n-something|<\epsilon$, right?
Best Answer
$f_n$ converges pointwise to the $0$ function. It is trivial to see it for $x=0,1$ so let $x\in(0,1)$ be fixed and notice that $$f_n(x)=nx(1-x)^n=xe^{n\log(1-x)+\log n}.$$ As $x>0$ we know that $\log(1-x)$ is a negative number, so $n\log(1-x)+\log n\to-\infty$ as $n\to\infty$; i.e. $f_n(x)\to0$ as $n\to\infty.$
So we know that $f_n$ converges pointwise to $0$ as $n\to\infty$. To see that this convergence is not uniform let's first find the maximum of $f_n$. This polynomial is positive in $(0,1)$ and $0$ in $x=0,1$ so it must attain its maximum at some $x_n\in(0,1)$. As $\log x$ is an increasing function we can maximize $\log f_n$ instead of $f_n$. As $$ \frac{d}{dx}\log f_n(x)=\frac{d}{dx}\left[\log n+\log x+n\log(1-x)\right]=\frac{1}{x}-\frac{n}{1-x}=\frac{1-x-nx}{x(1-x)} $$ equals zero only for $x=\frac{1}{1+n}$, this is the only critic point of $f_n$ and therefore $x_n=\frac{1}{1+n}$. Now we are done because \begin{align} \sup_{0\leq x\leq1}f_n(x)=f_n(x_n)=&\frac{n}{1+n}\left(1-\frac{1}{1+n}\right)^n\\ =&\frac{n}{1+n}\underbrace{\left(1-\frac{1}{1+n}\right)^{-1}}_{=(1+n)/n}\left(1+\frac{-1}{1+n}\right)^{1+n}\underset{n\to\infty}{\to}e^{-1}>0. \end{align}