$f:\mathbb{R} \rightarrow \mathbb{R}$ with $f(0)=0$ and $|f'| \leq |f|$ then $f$ is constant

Question

If $f:\mathbb{R} \rightarrow \mathbb{R}$ is differentiable with $f(0)=0$ and $|f'| \leq |f|$, then $f$ is constant in $[0,1]$. Trying to use the mean value theorem, but I am stuck.

Answer 1

Let $M = \max\{|f(x)| : 0 \le x \le 1\}$ and let $x_0 = \min \{x : |f(x)| = M \}$.

If $x_0 = 0$ then $M = |f(0)| = 0$ which implies $f \equiv 0$.

If $x_0 > 0$ then by the mean value theorem there exists $c \in (0, x_0)$ such that

$$ |f(x_0)| = |f(x_0) - f(0)| = |f'(c)||x_0 - 0| \le |f(c)| $$

since $x_0 \le 1$ and $|f'(c)| \le |f(c)|$. But $c < x_0$ means that $|f(c)| < M$ which gives us $M = |f(x_0)| < M$, a contradiction.

Answer 2

Let $x_0$ be a point in $[0,1)$ where $f(x_0)\neq 0.$

Then let $x_1$ be a point in $[0,x_0]$ where $|f(x)|$ reaches the maximum on that interval.

But then, by the mean value theorem

$$f(x_1)=f(x_1)-f(0)=f'(c)x_1$$

For some $c\in(0,x_1)$.

Then, since $0<x_1<1$, and $f(x_1)\neq 0$, you have that $|f(x_1)|<|f'(c)|<|f(c)|$, which contradicts our definition of $x_1$.

Answer 3

Let $P(n)$ be : $\forall x\in (0,1]\; \exists x_n\in (0,x)\; (|f(x)\leq |xf(x_n)|.$

For $x\in (0,1]$ there exists $x_1\in (0,x)$ such that $$\left|\frac {f(x)}{x}\right|=\left|\frac {f(x)-f(0)}{x-0}\right|=|f'(x_1)|\leq |f(x_1)|$$ implying $P(1).$

If $P(n)$ for some $n\in \mathbb N$ then for all $x\in (0,1]$ we have $|f(x)\leq |x^n f(x_n)|$ for some $x_n\in (0,x).$ Applying $P(1)$ to $x_n ,$ there exists $(x_n)_1\in (0,x_n)$ such that $|f(x_n)|\leq |x_nf((x_n)_1))|.$ Hence $$|f(x)|\leq |x^n\cdot x_nf((x_n)_1)|\leq |x^{n+1}f((x_n)_1)|.$$ Now $(x_n)_1\in (0,x)$ so let $x_{n+1}=(x_n)_1,$ and we obtain $$P(n)\implies P(n+1).$$ By induction on $n $ we have $$\forall n\in \mathbb N\;(P(n)).$$ Let $M=\sup \{|f(y)|:y\in [0,1]\}.$ Since $f$ is differentiable, $f$ is continuous, so $M<\infty.$ Now for all $x\in (0,1)$ there exists $x'_n\in (0,x)$ such that $$|f(x)|\leq |x^nf(x_n)|\leq x^nM.$$ So $f(x)=0$ for all $x\in (0,1).$

Answer 4

By continuity at $0$, for any $\varepsilon\gt0$, there exists $\delta\gt0$ such that for any $x$, $\lvert x\rvert\le\delta$ implies $\lvert f(x)\rvert\le\varepsilon$. Note also that $\lvert f'(x)\rvert\le\varepsilon$ for these $x$.

Now, by mean value theorem, for any $x$ with $0\lt\lvert x\rvert\le\delta$, $\big\lvert\frac{f(x)}{x}\big\rvert=\lvert f'(c)\rvert\le\varepsilon$ for some point $c$. Hence, $\lvert f(x)\rvert\le\varepsilon\lvert x\rvert\le\varepsilon\delta$ for any $x$ with $0\lt\lvert x\rvert\le\delta$. Then we have $\lvert f'(x)\rvert\le\varepsilon\delta$ for all $x$ with $0\lt\lvert x\rvert\le\delta$. The above argument is repeated to obtain $\lvert f(x)\rvert\le\varepsilon\delta^n$ for each $n\in\Bbb N$ and any $x$ with $0\lt\lvert x\rvert\le\delta$. See what happens when $\delta\lt1$.

Now choose $\varepsilon=\max\{\lvert f(x)\rvert:-1\le x\le1\}+1$ (this extra $+1$ is to prevent $\varepsilon$ to be $0$) so that $\delta$ can be any number in the interval $(0,1)$, i.e. we obtain $f$ is $0$ on $(-\delta,\delta)$ for each $\delta\in(0,1)$. Do NOT choose $\delta=1$ because the above inequality $\lvert f(x)\rvert\le\varepsilon\delta^n$ will not force $f(x)$ to be $0$ when $\delta=1$. Extend by continuity to see $f(1)=0$ as well.

Answer 5

If we define $g:[0,2]\to \mathbb{R}$ through $$ g(x)=\left\{\begin{array}{rcl} f(x) &\text{if}& x\in[0,1]\\ f(2-x)&\text{if}& x\in[1,2]\end{array}\right.$$ we are in the hypothesis of the second version of Wirtinger's inequality for functions, giving:

$$ 2\int_{0}^{1}f(x)^2\,dx = \int_{0}^{2}g(x)^2\,dx \leq \frac{4}{\pi^2}\int_{0}^{2}g'(x)^2\,dx =\frac{8}{\pi^2}\int_{0}^{1}f'(x)^2\,dx$$ but since $|f'|\leq|f|$ we have $$ \int_{0}^{1}f(x)^2\,dx \leq \frac{4}{\pi^2}\int_{0}^{1}f'(x)^2\,dx \leq \frac{4}{\pi^2}\int_{0}^{1}f(x)^2\,dx $$ hence $\int_{0}^{1}f(x)^2\,dx = 0$. Since $f\in C^1$, $f(x)$ equals zero for any $x\in[0,1]$.