# testing uniform convergence on [0,1]

hi everyone i'm having a couple of issues determing the convergence of two functions. normally i would look to prove that they converge pointwise then test for uniform convergence.

so a function converges pointwise if $\forall \epsilon > 0 \text{ and } \forall x\in A \text{ there exists } N\in \mathbb{N} \text{ so that when } n>N \text{ we have } |f_n(x)-f(x)| \leq \epsilon$

and a function converges uniformly if $\forall \epsilon > 0, \text{ there exists a } N\in \mathbb{N} \text{ such that when } n>N \text{ implies } \sup_{x \in A}|f_n(x)-f(x)| \leq \epsilon$

so im testing for uniform convergence for the function $$f_n(x) = \frac{x^n}{1+x^n}$$ for $x \in [0,1]$

my gut instinct when looking at this is tells me this will converge to zero unfortuantly stating "this will converge to zero because i think it will" doesnt suffice. i've tried differenting $f_n$ and i find that $f'_n=0 \text{ at } x=0$ the second derivative test tells me this is a point of inflection so i'm unsure how to go about this.

once i know the limit is zero though i can show this doesnt converge uniformly on the inteval as $$\sup_{x \in [0,1]} |f_n(x)-f(x)|=\sup_{x \in [0,1]} |f_n(x)|=\sup_{x \in [0,1]} |\frac{x^n}{1+x^n}|=\frac{1}{2}$$

any and all guideance would be great. thanks

There is a common trick for these, based on the fact that the uniform limit of continuous functions is continuous. Here you have $$\lim_{n\to \infty}f_n(x)=0$$ if $x\in [0,1)$. But,

$$\lim_{n\to \infty}f_n(1)=1/2$$ can you conclude from here?

• so is it for the fact that at n large enough we have a jump from 0 to 0.5? making the limit discontinuous at 1? which when looking at demos grapher looks around n=100000..... – johnnyB Oct 5 '17 at 2:03
• it's that the limiting function is 0 everywhere except at 1, where it's a 1/2. Thus it is not continuous – qbert Oct 5 '17 at 2:32

Another approach which shows that convergence is not uniform even if we restrict the domain to $[0,1)$:

It's easy to check that $f_n(x) \to 0$ pointwise on $[0,1)$.

On the other hand, \begin{aligned} \frac{x^n}{1+x^n} > \frac{1}{3} &\iff 3x^n > 1 + x^n \\ &\iff 2x^n > 1 \\ &\iff x^n > \frac{1}{2} \\ &\iff x > \frac{1}{2^{1/n}} \end{aligned}

so in particular, $$\sup_{x \in [0,1)} f_n(x) > \frac{1}{3}$$ for all $n$.

• so where as qbert is using weierstrass' theorem youre showing $sup f_n$ is always larger than 1/3. (sorry its 3am here my mind has slowed now) and this implies pointwise only because the two limits differ? – johnnyB Oct 5 '17 at 2:10