# Exercise on derivatives from Rudin's Principles of Mathematical Analysis

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 $\beta.\;\square$

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

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I've edited your post a bit. For future reference, don't do your own formatting with $\quad$s. The spacing may look okay on your computer, but on most other people's web browsers with different font sizes and screen widths, the output would look absolutely horrible. For displayed equation use  to enclose the math expression, and use \tag to get equation number/symbols. – Willie Wong Dec 18 '12 at 12:52
Oh, the inequality is supposed to go the other way. I'll edit it. – docholliday Dec 18 '12 at 13:16

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 $g(-a) = y_0 > 0$.