Exercise on derivatives from Rudin's Principles of Mathematical Analysis

Question

Exercise 5.26 --Rudin [Principle of Mathematical Analysis]

Ex.5.26 If $f'$ exists on $[a,b]$, $f(a)=0$ and $\exists A\in\mathbb{R}\;(|f'|\le A|f|\,\text{on }[a,b])$, then $f = 0$ on $[a,b]$.

Hint: Fix $x_0\in [a,b]$, let $M_0 = \sup |f|([a,x_0])$, $M_1=\sup |f'|([a,x_0])$. Then $$|f(x)|\le M_1(x_0 -a)\le A(x_0-a)M_0$$ (for $x\in [a,x_0]$). Hence $M_0= 0$ so $f = 0$ on $[a,x_0]$ if $A(x_0 - a) < 1$. Proceed.

Now I've done the above but I'm asked to assume $f(a) = y_0 > 0$. Show that $f(t)\le y_0*e^{A(t-a)}$.

How do I take care of the case where $f(t) = 0$ for some $t > a$? I'm then asked to examine $\ln (f(t))$?

Proof for 5.26: Assume $A>0$ (otherwise nothing to prove) and let $\beta = \sup \{c\in [a,b]: f([a,c]) = \{0\}\}$. Then $\beta \in [a,b],\; f([a,\beta]) = \{0\}$ since $f(a)=0$ and $f$ is continuous. we shall show $\beta = b$.

If $\beta < b$, let $\gamma = \min(b,\beta+\frac{1}{A})$ and take $\beta_1\in (\beta, \gamma)$ then for $x\in [\beta,\beta_1]$ we have $|f(x)| = |f(x) - f(\beta)|\le M_1(x-\beta)\le A(\beta_1-\beta)M_0$ and $A(\beta_1-\beta) < A(\gamma-\beta)\le 1$ Where $M_0 = \sup |f|([\beta,\beta_1]),\; M_1 = \sup |f'|([\beta,\beta_1])$. Thus $M_0 \overset{(\star)}{=} 0$ but then we get $f([a,\beta_1]) = f([a,\beta]\cup [\beta,\beta_1]) = \{0\}$ and $\beta < \beta_1 \in [a,b]$, contradicts to the def. of $\beta.\;\square$

$$(0\le M_0 \le sM_0 \wedge s<1)\implies (0\le (1-s)M_0 \le 0)\implies (M_0=0)\tag{$\star$}$$

Answer 1

For your specific question of how to rule out $f(t) = 0$ for $t > a$, one way to do it is by contradiction using Ex 5.26 itself. Suppose that for $c \in (a,b]$ we have that $f(c) = 0$. Let $g(t) = f(-t)$. Then we have still $|g'(t)| \leq A |g(t)|$ since the same is true for $f$. We have that $g$ is differentiable in $[-c,-a]$ with $g(-c) = 0$. So applying Ex 5.26 $g\equiv 0$, contradicting that $g(-a) = y_0 > 0$.