A sequence of continuous functions $(f_n\colon[a,b]\to\mathbf{R})_{n}$ is said to be point-wise bounded if for all $x\in[a,b]$ there is a $R_x>0$ such that $$|f_n(x)|\le R_x\quad\mbox{for all }n.$$

How can I visualize this definition? What kind of picture should I associate with this definition?

Thanks in advance.

  • 2
    $\begingroup$ I edited the tags of this post to include "intuition". $\endgroup$ – Kirk Boyer Aug 31 '12 at 23:16

Short answer: if $(f_{n}:[a,b]\rightarrow\mathbb{R})_{n}$ is pointwise-bounded, then there is another function $g:[a,b]\rightarrow\mathbb{R}$ such that $|f_{n}(x)| \leq g(x)$ for all $x$ and all $n$. Visually, this means that if you draw the graph of $g$ and the graph of $-g$, then the graphs of all the $f_{n}$ will be between them.

Visualizing things like this sometimes becomes easier when you know what a counterexample looks like. What kinds of sequences of functions are $not$ pointwise-bounded?

An easy example is given by $f_{n}(x) = n$, the sequence of natural-number constant functions. This sequence breaks pointwise boundedness at $every$ point in the reals, so it doesn't matter what $[a,b]$ is; fix any real number $x$ and the value of $f_{n}$ is unbounded as $n$ increases.

Now, consider the sequence of functions given by $f_{n}(x) = x^{n}$. If $x_{0}\in [0,1]$, then it doesn't matter how big you make the exponent; $x_{0}^{n}$ will also be between $0$ and $1$. But if $x_{1}>1$, for example, then the sequence of function "blows up"; you can make the value of $|f_{n}(x_{1})|$ as big as you want by making $n$ big enough. In this case, we say the sequence $f_{n}:[0,1]\rightarrow \mathbb{R}$ defined this way is pointwise bounded, whereas (for example) $f_{n}(x):[0,2]\rightarrow \mathbb{R}$ is not pointwise bounded, since fixing x anywhere above 1 makes it so you can't pick a bound for the function values.

Often when we talk about "pointwise" properties, we do so because we want to contrast this with the "uniform" counterpart; in this case, uniform boundedness. Visualizing uniform boundedness is actually easier to do (which is different from the case for other many uniform properties): if a sequence of functions is uniformly bounded, there is some bounding number $B>0$ such that regardless of our choice of $x$ in the domain, we have that $|f_{n}(x)|\leq B$. To characterize this in the same terms as the pointwise situation, we still have a function $g(x)$ whose graph envelops all the graphs of the $f_{n}$'s, but this time $g$ is a constant function.


Define a new function $r:[a,b]\to\Bbb R:x\mapsto R_x$. The condition that $|f_n(x)|\le R_x$ for all $n$ and all $x\in[a,b]$ just says that for each $n$, the graph of $y=f_n(x)$ lies between the graphs of $y=r(x)$ and $y=-r(x)$. Of course these two graphs could be very ugly, because the bounding function $r(x)$ might be wildly discontinuous, but they do give you an irregular ‘strip’, symmetric about the $x$-axis, in which all of the graphs $y=f_n(x)$ must lie.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.