Showing if a function is uniformly convergent.

Question

I got the series of functions $f_n=x^n(1-x)$ on $[0,1]$. It does converge pointwise to $f$, $f(x)=0$. It looks like it's uniformly convergent, but I've got this example that shows it's not:

$||f_n-f||_{sup} \geq$ $ sup_{x \in [0,1]}|f_n((1/2)^{1/n})-f((1/2)^{1/n})|$ =$\geq lim_{x\rightarrow0}=1/2 - 1/2x$=$1/2$ not zero. I know the writing isn't precise but I just wanna know why $(1/2)^{1/n}$ doesn't work?

What is wrong about my reasoning? I showed with (1/2)^1/n, that the convergence is not uniform. But it is. — Jack, May 18 '18 at 16:38
You have: $$f_n ((1/2)^\frac{1}{n})=1/2 (1-(1/2)^\frac{1}{n})$$ and $$\lim_{n \to \infty} (1/2)^\frac{1}{n} =1$$ so the limit of $f_n ((1/2)^\frac{1}{n})$ is $0$. — Delta-u, May 18 '18 at 16:46
In fact $f_n$ does converge uniformly, see here: https://math.stackexchange.com/questions/762777/how-to-prove-uniform-convergence-for-sequences-f-n-xn1-x-f-n-frac-n3 — Math1000, May 18 '18 at 17:48

score 1 · Answer 1 · answered May 18 '18 at 21:35

You can use derivatives to guide you in proving whether the sequence converges uniformly or not: Let $n$ be given. Then $$f_n'(x)=nx^{n-1}(1-x)-x^n=x^{n-1}(n-nx-x)$$ so $f_n'(x)=0$ iff $x=0$ or $x=n/(1+n)=1-\frac{1}{n+1}$. This means that $f_n$ attains the maximum at $x_n=1-\frac{1}{n+1}$, and in this case \begin{align*} \Vert f_n\Vert_\infty=f_n(x_n)=\left(\frac{n}{1+n}\right)^n\frac{1}{n+1}\leq\frac{1}{n+1} \end{align*} so $f_n$ does converge uniformly to $0$.

Showing if a function is uniformly convergent.

1 Answers1