15
$\begingroup$

This question pertains to the sequence of functions $f_n(x)=x^n$ on the interval $[0,1]$.

It can be shown this sequence of functions ${f_n}$ converges point-wise to the limit $f$ where $f$ is defined by $f(x)=0$ on $[0,1)$ and $f(x)=1$ at $x=1$.

However, this sequence of functions ${f_n}$ does not converge uniformly to $f$. One way to prove this (which I have seen) is via a theorem which proves that if a sequence of functions ${f_n}$ converges uniformly to $f$, then $f$ is continuous. And clearly it is not the case that $f$ is continuous in our example, so our convergence is not uniform.

However, I have seen another test for uniform convergence on an interval $s$. That is that:

$${\rm lim}\ [{\rm sup}\ \{ | f_n(x)-f(x)|\ :\ x\in S\ \}]=0$$

This conception of uniform convergence can be found in my Walter Rudin analysis book for example. My question is, how can we use this definition of uniform convergence to show $x^n$ is not uniformly convergent?

I am having trouble seeing how to apply this definition to this example, especially since the limit $f$ is defined piece-wise. But I know I need to find the sup of the difference between $f_n$ and $f$ over the interval $[0,1]$ (how do I find this sup?). And then once I find it, I need to take the limit as $n\to\infty$ and show it does not equal $0$.

Thank you for your help!

$\endgroup$
1
  • $\begingroup$ You probably want something explicit. Pick $\epsilon<1$ small. Check that for any $\delta>0$ there is an $N$ (you can explicitly find a formula for $N$ in terms of $\epsilon,\delta$) such that if $n>N$, then $|f_n(x)-f(x)|<\epsilon$ if $0\le x\le 1-\delta$. Use this to explicitly verify that uniform convergence is violated. $\endgroup$ Commented Jan 27, 2014 at 2:19

3 Answers 3

20
$\begingroup$

It all boils down to proving that $$\sup_{x\in [0,1]}|x^n-\chi_{\{1\}}|=\sup_{x\in [0,1)}|x^n-0|=\sup_{x\in[0,1)}x^n=1\not\to 0$$

$\endgroup$
4
  • 4
    $\begingroup$ @Pedro Tamaroff: What is $\chi_{\{1\}}$? $\endgroup$ Commented Jul 7, 2016 at 14:54
  • $\begingroup$ It is the indicator function for $x=1$. $\endgroup$
    – Brofessor
    Commented Sep 5, 2017 at 13:45
  • $\begingroup$ Can you explain, why you can go from $[0,1]$ to $[0,1)$? $\endgroup$
    – mdcq
    Commented Jan 8, 2018 at 17:51
  • $\begingroup$ @philmcole it's because $|f_n(1) - f(1)| = 0 $ so since supremum is $\geq 0$, we can exclude that point $\endgroup$
    – chesslad
    Commented Nov 5, 2019 at 18:22
5
$\begingroup$

I think the best way to see that this function doesn't converge uniformly on $x \in [0,1]$ is to note that the limiting function is discontinuous for $x \in [0,a], a<1$ and $x=1$: $$ \lim_{n \to \infty} f_n(x)= \left\{ \begin{array}{rl} 0 &x \in [0,a]\\ 1 & x=1 \end{array} \right. $$

$\endgroup$
0
$\begingroup$

Let's recall the definition of uniform convergence:

$$ f_n (x) \rightarrow f(x) \text{ uniformly on $\Delta$ if:} \\ \forall \epsilon > 0, \exists N_{\epsilon} \in \mathbb{N} : \forall n \geq N_{\epsilon} \implies |f(x) - f_n(x)| < \epsilon, \forall x \in \Delta \tag{1} $$

In order to show that the convergence is non-uniform, we can either apply the Weierstrass M-test, or we can consider the converse of (1), by picking some $x_0$ in a clever way such that we can bound $|f(x) - f_n(x_0)|$ below by a positive constant (or in practice, we often choose $x_0$ such that $|f(x) - f_n(x_0)|$ is constant)

In this case, we can take $x_0 = a^{\frac{1}{n}}$, where $a \in [0, 1)$ and we can clearly see that the pointwise limit $x^n$ is 0

Thus $|f(x) - f_n(x)| = (a^{\frac{1}{n}})^{n}= a$, which does not converge to 0 (i.e. this difference is constant, thus can't be made arbitrarily small), hence the convergence is non-uniform.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .