# Counterexample around Dini's Theorem

"Give an example of an increasing sequence $$(f_n)$$ of bounded continuous functions from $$(0,1]$$ to $$\mathbb{R}$$ which converge pointwise but not uniformly to a bounded continuous function $$f$$ and explain why Dini's Theorem does not apply in this case"

So clearly Dini's Theorem does not apply, as $$(0,1]$$ is not a closed interval (or compact metric space), but I can't figure out an example.

My first thought is $$f_n(x)=\frac{1}{x^n}$$, but this does not converge pointwise to a bounded continuous function, as $$x=1$$ is in the interval

My second thought is $$f_n(x)=x^\frac{1}{n}$$. This is clearly an increasing sequence of bounded continuous functions (I think?) I believe this converges pointwise to $$f(x)=1$$ for all $$x\in (0,1]$$, but I'm struggling to then show why this doesn't converge uniformly to $$f(x)=1$$

How would I do this? Or is then an easier/better example I could use?

• Regarding your second thought - pick an $\epsilon$, say $\frac{1}{2}$. Now for a given $n$, can you always find an $x$ so that $x^{\frac{1}{n}} < \frac{1}{2}$? If so then you're basically done. – MartianInvader Nov 20 at 17:44

Take $$f_n(x)=e^{-\frac 1 {nx}}$$ and $$f=1$$.
Note that $$sup_x |f_n(x)-f(x)| \geq |f_n(\frac 1 n)-1|=1-\frac 1 e$$.
Try $$f_{n}(x)=1-(1-x)^{n}$$ for $$x\in(0,1]$$ and $$f=1$$.
As a sequence where it is easy to reason with, you could also use $$f_n(x) = \min(nx, 1)$$ and $$f = 1$$.
Your example $$f_n(x) = x^{1/n}$$ works fine. What you need is to show that for every $$n$$ there is an $$x_n$$ such that $$|f_n(x_n)-f(x_n)|$$ is greater than some positive constant, for example $$x_n = 1/2^n$$. That way $$f_n$$ cannot converge uniformly to $$f$$.