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"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?

Klangen
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Dino
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  • 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 '19 at 17:44

4 Answers4

7

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 $.

Sine of the Time
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Kavi Rama Murthy
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6

Try $f_{n}(x)=1-(1-x)^{n}$ for $x\in(0,1]$ and $f=1$.

user284331
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3

As a sequence where it is easy to reason with, you could also use $f_n(x) = \min(nx, 1)$ and $f = 1$.

Carsten S
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Jens
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3

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 $f$.

Magma
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