I have already proven the following property (taken from Baby Rudin Chapter 5 Exercise 26), which I think is supposed to help me in some way:
If $f$ is differentiable on $[a, b]$, $f(a) = 0$, and there is a real number $A$ such that $|f^\prime x)| \leq A|f(x)|$ on $[a, b]$, then $f(x) = 0$ for all $x \in [a, b]$.
I am trying to prove the following properties about the exponential function $\exp : \mathbb{R} \rightarrow \mathbb{R} $ using the definition
$$
[\exp(x)]^\prime = \exp(x) \\
\exp(0) = 1.
$$
(I'll provide some brief sketches of ideas I have, and I would greatly appreciate if someone could point out the flaws in my reasoning or provide some suggestions to create better arguments.)
- $\exp(x) > 0$ and $\exp(x)$ is a strictly increasing function of $x$.
It seems that we can proceed by showing that $\exp(x)$ can never change sign. If it did, i.e. $\exp(a) = 0$ for some $a \in \mathbb{R}$, then, since we have $$ |[\exp(x)]^\prime| = |\exp(x)| \leq A |\exp(x)|, $$ we would have that $f(x) = 0$ for all $x \in [a, \infty]$. However, I'm not sure what to do about the case where $a > 0$ because then it doesn't contradict $\exp(0) = 1$.
In any case, after showing that $\exp(x) > 0$, we then get $[\exp(x)]^\prime > 0$ for free, which shows that $\exp$ is strictly increasing by Rudin Theorem 5.11.
- $\exp(x+y) = \exp(x)\exp(y)$ for all $x, y \in \mathbb{R}$.
I took the derivative of $$ f(x) = \exp(x+y) - \exp(x)\exp(y), $$ using properties of derivatives listed in Rudin: $$ \begin{align*} f^\prime (x) &= [\exp(x+y) - \exp(x)\exp(y)]^\prime \\ &= [\exp(x+y)]^\prime - [\exp(x)\exp(y)]^\prime \\ &= \exp(x+y) - \exp(y)[\exp(x)]^\prime \\ &= \exp(x+y) - \exp(x)\exp(y) \end{align*} $$
This again means that $|f^\prime (x)| \leq A|f(x)|$, and clearly $f(0) = 0$, so by the exercise we have that $\exp(x+y) = \exp(x)\exp(y)$ for all $x \in [0, \infty]$, $y \in \mathbb{R}$. Perhaps we could repeat the argument in terms of $y$ to get the $x \in \mathbb{R}$ part?
- $\exp(-x) = \frac{1}{\exp(x)}$ for all $x \in \mathbb{R}$.
From (2), $$ \exp(0) = 1 = \exp(x - x) = \exp(x)\exp(-x), $$ and the desired result follows.
Thanks!
(1). If $g'=g$ and $g(x_0)=0$ for some $x_0$ then $g(x)=0$ for all $x.$
Proof: Case 1.If $[-1/2+x,1/2+x]\subset g^{-1}\{0\}$ for all $x\in g^{-1}\{0\}$ then $\Bbb R=\cup_{n\in \Bbb N}[-n/2+x_0,n/2+x_0]\subset g^{-1}\{0\}.$
(ii). Case 2. Suppose $g(x)=0$ and $|y-x|\le 1/2$ and $g(y)\ne 0.$ Then since $g$ is continuous there exists $z$ with $|z-x|\le |y-x|$ and $g(z)=g(y)$ and such that $|g(w)|\le|g(y)|$ for all $w$ between $x$ and $z.$ Now $g'$ is continuous (as it is equal to the differentiable, hence continuous, $g$) so $$0\ne |g(z)|=|g(z)-g(x)|=|\int_x^zg'(w)dw\;|=$$ $$=|\int_x^z g(w)dw\;|\le$$ $$\le |\int_x^z|g(w)|dw\;|\le $$ $$\le |\int_x^z |g(z)|dw\;|=$$ $$=|z-x|\cdot |g(z)|\le |g(z)|/2$$ which is absurd.
(2). If $f'=f$ and $f(0)=1$ then $f$ is strictly increasing and positive.
Proof: By (1) we cannot have $f(x)=0$ for any $x.$ And if $f(x)<0$ for some $x,$ then since $f$ is continuous and $f(0)>0,$ there would be some $y$ between $x$ and $0$ with $f(y)=0.$ So $f$ is positive. Now since $f' (=f)$ is positive, $f$ is strictly increasing
(3). With $f$ as in (2), for any $y$ let $g_y(x)=f(x)f(y)-f(x+y).$ Then $\frac {dg_y(x)}{dx}=g_y(x)$ and $g_y(0)=0.$ By (1), $g_y(x)=0$ for all $x.$ This holds for any $y,$ so $f(x)f(y)-f(x+y)=0$ for all $x,y.$
(4). It remains to show that there $does$ exist $f$ with $f'=f$ and $f(0)=1.$