# Prob. 26, Chap. 5 in Baby Rudin: If $\left| f^\prime(x) \right| \leq A \left| f(x) \right|$ on $[a, b]$, then $f = 0$

Here is Prob. 26, Chap. 5, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ is differentiable on $[a, b]$, $f(a) = 0$, and there is a real number $A$ such that $\left| f^\prime(x) \right| \leq A \left| f(x) \right|$ on $[a, b]$. Prove that $f(x) = 0$ for all $x \in [a, b]$. Hint: Fix $x_0 \in [a, b]$, let $$M_0 = \sup \left| f(x) \right|, \qquad M_1 = \sup \left| f^\prime(x) \right|$$ for $a \leq x \leq x_0$. For any such $x$, $$\left| f(x) \right| \leq M_1 \left( x_0 - a \right) \leq A \left( x_0 - a \right) M_0.$$ Hence $M_0 = 0$ if $A \left( x_0 - a \right) < 1$. That is, $f = 0$ on $\left[ a, x_0 \right]$. Proceed.

My Effort:

Let $n$ be a positive integer such that $n > A(b-a)$, and let $a = x_0 < x_1 < \cdots < x_n = b$ be real numbers such that $x_i = a + \frac{i(b-a)}{n}$ for $i = 0, 1, \ldots, n$. Thus we have partitioned the closed interval $[a, b]$ into the $n$ sub-intervals $\left[ a, a + \frac{b-a}{n} \right]$, $\left[ a + \frac{b-a}{n}, a + \frac{2(b-a)}{n} \right]$, $\ldots$, $\left[ a + \frac{(n-1)(b-a)}{n}, b \right]$, such that the length of each sub-interval is $\frac{b-a}{n} < {1 \over A}$.

As $\left| f^\prime (x) \right| \leq A \left| f(x) \right|$ on $[a, b]$, so for any $x \in \left[ a, x_1 \right]$, we have $$\left| f^\prime (x) \right| \leq A \left| f(x) \right| \leq A M_0,$$ where $$M_0 = \sup \left\{ \ \left| f(x) \right| \ \colon \ a \leq x \leq x_1 \ \right\}, \tag{0}$$ which implies that $$M_1 = \sup \left\{ \ \left| f^\prime (x) \right| \ \colon \ a \leq x \leq x_1 \ \right\} \leq A M_0. \tag{1}$$

Now as $f$ is differentiable on $\left[ a, x_1 \right]$, so for each $x \in \left( a, x_1 \right]$, $f$ is continuous on $\left[ a, x \right]$ and differentiable on $\left( a, x\right)$. Therefore by the mean-value theorem there is a point $c_1 \in \left( a, x \right)$ such that $$f(x) = f(a) + f^\prime\left( c_1 \right) \left( x-a \right) = f^\prime\left( c_1 \right) \left( x-a \right),$$ and so if $M_0 > 0$, then $$\left| f(x) \right| = \left| f^\prime\left( c_1 \right) \right| \left( x-a \right) \leq M_1 \left( x_1 - a \right).$$ Thus the real number $M_1 \left( x_1 - a \right)$ is an upper bound for the set in (0) above, so we must have $$M_0 \leq M_1 \left( x_1 - a \right) \leq A M_0 \left( x_1 - a \right) = A M_0 \left( \frac{b-a}{n} \right) < M_0 ,$$ which gives rise to a contradiction. So $M_0 = 0$. Therefore $f(x) = 0$ for all $x \in \left[ a, x_1 \right]$.

Now applying a similar argument to each of the remaining sub-intervals we can conclude that $f(x) = 0$ for all $x \in [a, b]$.

Is this proof correct and as Rudin has required? If not, then what have I missed?

• My hint would have been $\frac{f'(x)}{f(x)}=(\log|f(x)|)'$ when $f(x)\neq 0$. Commented Jun 9, 2017 at 20:42
• @ThomasAndrews we have not yet discussed the logarithmic function in a rigorous enough manner. Commented Jun 9, 2017 at 20:51
• This proof is indeed correct. Commented Jun 10, 2017 at 1:17

Let $c$ be such that $c\leq b$ and $f(x)\neq 0$ for all $x\in (a,c]$, then by mean value theorem in $[a,x]$, there exist $a\leq\xi\leq x$ such that $$\left|\frac{f(x)-f(a)}{x-a}\right|=\left|\frac{f(x)}{x-a}\right|=\left|f'(\xi)\right|\leq A|f(\xi)|\textrm{, }\forall x: a<x\leq c.$$ Hence $$|f(x)|\leq |x-a|A|f(\xi)|\textrm{, }\forall x\in[a,c].\tag 1$$ Setting $x=\xi$ in (1) we get (since $a<\xi\leq c$ and $f(\xi)\neq 0$): $$|\xi-a|\geq\frac{1}{A}.\tag 2$$ But clearly $\xi$ could be near to $a$ as close as we want. This can be done by selecting another $c$ ''say'' $c'$ : $c'<c$. From the fact that $A$ is a fixed constant and (2) we lead to contradiction. QED