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Thanks to books, many pdf files on google and this web-site, I understood somewhat about uniform and pointwise convergence. This question may be the last question about this part.

Now, I can check whether the sequence of functions $f_k$ does not converge, converges pointwise, or converges uniformly by using the definitions.

However, I cannot check it, only depending on graph.


Let me give you some examples.

$$f_n(x)=x^n, ~~~~~ \mbox{for all}~~x \in [0, 1)$$

$\lim_{n\to\infty}f_n(x)=0$ because $|x|\lt1$. Hence, $\{f_n\}$ converges pointwise, whose pointwise-limit function $f$ is the zero function on the domain $D=[0, 1)$.

Now, I am checking whether the sequence $\{f_n\}$ converges uniformly or not.

for $\displaystyle\varepsilon=\frac{1}{3}$, $~~\nexists N\in\mathbb{R}$ such that $\displaystyle\left|f_n(1-\frac{1}{n})-f(1-\frac1n)\right|=\left(1-\frac1n\right)^n\lt\varepsilon~~~$ for every $n\gt{N}$ and for every $x\in{D}$.

The reason why $N$ does not exist is that even if $n$ approaches to $\infty$, $\displaystyle\left(1-\frac1n\right)^n=\frac{1}{e}\gt\varepsilon=\frac{1}{3}$.

Therefore, the sequence of functions $\{f_n\}$ converges pointwise but doesn't uniformly.


I tried to draw the functions by using Matlab.

enter image description here

However, I failed to find intuitively that it converges only pointwise only by seeing graph. Is there any methods to know $\{f_n\}$ converges uniformly or pointwise with only graph?

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    $\begingroup$ Just look at the picture-it doesn't matter how big you make $n$, for some $x$es, $f_n(x)$ is still far from $0$! $\endgroup$ – Kevin Carlson Mar 14 '16 at 0:58
  • $\begingroup$ Try comparing the plots you have to the plots of $g_n(x):=\frac{\sin(x)}{n}$, and see if you can see a difference. $\endgroup$ – snulty Mar 14 '16 at 1:18
  • $\begingroup$ Do the pictures here and here help? I searched in Google Images for uniformly convergent site:math.stackexchange.com and sup norm site:math.stackexchange.com. Maybe you will find some other relevant posts among the results. $\endgroup$ – Martin Sleziak May 6 '16 at 4:07
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Examples of sequence of functions that converge uniformly

enter image description here

In this case:

  1. Limit function must be continuous

  2. sequence of functions eventually converges to the function (lies in an $\epsilon$-tube) as $N \to \infty$

Examples of sequence of functions that converge pointwise but not uniformly

enter image description here

Note (b) has an error it should be $f_n(x) = \exp(-nx)$

In this case:

  1. The limit function do not have to be continuous

  2. The sequence of functions do not have to lie fully in an $\epsilon$-tube of the limit function as $N \to \infty$

The latter is because in the definition of pointwise convergence:

$\forall \epsilon > 0, \forall x \in [a,b], \exists N\in \mathbb{N}$ s.t. $\forall n\in \mathbb{N}, n \geq N, \|f_n - f\| < \epsilon$

We can pick our $N$ large enough to make the latter condition $\|f_n - f\| < \epsilon$ true for a particular $x$, but not for all $x$ at once.

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  • $\begingroup$ Many examples are helpful. Thank you. $\endgroup$ – Danny_Kim Mar 14 '16 at 5:01
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According to the definition, sequence $f_n \to 0$ uniformly on interval $I$ if for every $\epsilon > 0$ there is $N$ such that $|f_n(x)| < \epsilon$ for all $n > N$ and all $x \in I$.

Graphically, what this means is that if you take an arbitrarily narrow horizontal band around the $x$ axis, ($\{(x,y): |y| < \epsilon\}$), for large enough $n$ the graph of $f_n$ will lie in this band.

In your picture, the graph of each function $f_n$ will poke up out of the band for $x$ near $1$. Therefore the sequence does not converge uniformly to $0$.

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  • $\begingroup$ Or, in otherwords, do the $\epsilon$-neighborhoods of each point converge to an $\epsilon$-neighborhood of the corresponding limit point? Notice how that fails to happen for $x=1$ in the given example. $\endgroup$ – Justin Benfield Mar 14 '16 at 1:03

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