# Prove that $f(x)$ is bounded between $e^x$ and $e^{-x}$

So we finished studying chapter 5 of Rudin on differentiation (Mean value theorem, Taylor's theorem etc) and this was given as a homework problem:

Let $f(x)$ be continuously differentiable on $[0, \infty)$ such that $f$ satisfies $f'(x) = \cos(x^2)f(x)$ for all $x \geq 0$, with $f(0) = 1$. Prove that $e^{-x} \leq f(x) \leq e^x$ for all $x \geq 0$.

Clearly, $x = 0$ then the result is trivial. I tried to use Taylor's theorem to note that if $x > 0$, then there exists $x_1 \in (0,x)$ such that $f(x) = 1 + xf'(x_1) = 1 + x \cos(x_1^2)f(x_1)$. This is where I'm stuck, since I don't know what to do with the cosine function. Any hint/help/comment is greatly appreciated.

• Take $g(x)=e^x$ and try to use the Mean value theorem $$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(x)}{g'(x)}$$ – Arun Feb 8 '17 at 0:41

$cos(x^2)\in [-1,1]$, so $f'(x)\geq-f(x)$ and $f'(x) \leq f(x)$. Multiplying the first relation with $e^x$ and the second with $e^{-x}$ you get $(f(x)e^x)' \geq 0$ and ($f(x)e^{-x})'\leq 0$, so $f(x)e^x$ is increasing and $f(x)e^{-x}$ is decreasing.
So $f(x)e^x \geq f(0)e^0=1$, hence $f(x) \geq e^{-x}$.
Also, $f(x)e^{-x} \leq 1$, hence $f(x) \leq e^x$.

• That is very interesting! Thank you. So the only property of the cosine function that we used was that it's between -1 and 1. The $x^2$ kinda threw me off. – user228960 Feb 8 '17 at 2:48

For all real x, $-1 \leq$cos$(x) \leq 1$

so $0 \leq$cos$^2 (x) \leq 1 \ \forall x \in R$

Given $f'(x) =$cos$^2 x \ f(x) \ \ \forall x \geq 0$, then

$|f'(x)| \leq |f(x)| \ \ \forall x\geq 0$

We know if $f'(x) = f(x)$on$\ R$ with $f(0) = 1$, then $f(x) = e^x$

You should be able to complete the proof from those two facts.

I assume $f\geq 0$ Hint: write $h(x)=e^{-x}f(x), h'(x)=e^{-x}f(x)(cos(x^2)-1)$

and $l(x)=e^{x}f(x)$, if $f\geq 0$, $h$ decreases and $l$ increases, so $h(x)\leq h(0)=1$ and $l(x)\geq l(0)=1$.