How to prove boundedness of a function on $\mathbb{R}$?

Question

I'm somewhat new to real analysis. I've seen proofs for boundedness on an interval. How do I prove boundedness on $\mathbb{R}$?

Say, how to prove that the sigmoid function $S$ is bounded? $S(x)=\frac{1}{1+e^{-x}}$?

Answer 1

That depends on the function in question. In your example it is easy to see that, for every $x$,

1. $S(x)>0$
2. $1+e^{-x} > 1 \Rightarrow \frac{1}{1+e^{-x}}<1$ So $S(x) $ is bounded from above and below independently of $x$.

A general rule does not exist, but often a succesful approach is to choose a bounded closed interval $[-R,R]$ for some $R>0$, and then investigate the behavior on $(-\infty, -R]$, $[R,R]$, and $[R,\infty)$ separately.

If you know for example, that a function is continuous on $[-R,R]$ it will be also bounded on $[-R, R]$. If it has a finite limit when $x\rightarrow \infty$, then (by definition of the limit) it will be bounded on some set $[R_0, \infty)$ and so on.

Answer 2

Try to look at the expression $1+e^{-x}$

The element $e^{-x}$ is positive for every $x\in \Bbb{R}$.

also, The element $e^{-x}$ is strictly decreasing.

which means that for every $x_1<x_2$ we get $e^{-x_2}<e^{-x_1}$

and therefore $1+e^{-x_2}<1+e^{-x_1}$

so the inverse of that will be:

$$\frac{1}{1+e^{-x_2}}>\frac{1}{1+e^{-x_1}}$$

which means the expression is strictly increasing.

then we check for $S(x)$ values in $-\infty$ and $\infty$:

$$\lim_{x\to \infty}\frac{1}{1+e^{-x}} = \frac{1}{1+e^{-\infty}}=\frac{1}{1+0}=1$$

$$\lim_{x\to -\infty}\frac{1}{1+e^{-x}} = \frac{1}{1+e^{\infty}}=\frac{1}{1+\infty}=0$$

so we can see that the sigmoid $S(x)$ is bounded below by 0 and bounded above by 1