# visualisation of pointwise boundedness

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?

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I edited the tags of this post to include "intuition". –  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.

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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.

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