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Let $f:[0,\frac 1 2]\to \mathbb R$ be differentiable and let $|f'(x)|\le|f(x)|$ and $f(0)=0$.

Prove that $\forall x\in[0,\frac 1 2]:f(x)=0$

I got stuck when I tried to solve this.

If we'll define a function: $g(x)=\ln(f(x))$ then $g'(x)=\frac {f'(x)}{f(x)}\le \frac {f(x)}{f(x)}=1$ so this function's derivative is bounded so the function is uniformly continuous but I don't know how to use it.

Note: I can't use integrals.

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GinKin
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2 Answers2

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By the MVT, and recalling that $f(0)=0$, $$ |f(x)|=|f(x)-f(0)| = |f'(c)|x \leq \frac{1}{2} |f(c)| \leq \frac{1}{2}\|f\|_\infty, $$ where $0 < c < x$. Since $x \in [0,1/2]$ is arbitrary, $$ \|f\|_\infty \leq \frac{1}{2}\|f\|_\infty, $$ and thus $\|f\|_\infty =0$. Here I used the standard notation $\|f\|_\infty = \max_{0 \leq x \leq 1/2} |f(x)|$.

Siminore
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  • Why is it implied that $|f|\infty \leq \frac{1}{2}|f|\infty$ ? – GinKin Jun 24 '14 at 09:59
  • Because $|f(x)|$ is dominated by a quantity that is independent of $x$, and therefore also $\sup_x |f(x)|$ is dominated by the same quantity. In some sense, it is the definition of least upper bound. – Siminore Jun 24 '14 at 10:56
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Your definition of $g$ is not a good way to go since you know that $g(0)$ is undefined...

My advice is to focus more on the fact that $f(x)\approx x\cdot f'(0) + f(0)$.

5xum
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  • You suggest to use the Taylor expansion of $f$? So we'll get $f(x)= x\cdot f'(0) + \frac{f''(c)x^2}{2}$ for $c$ between $0$ and $x$. I don't see what to do next. – GinKin Jun 24 '14 at 09:36
  • You do not know that $f''$ even exists, so no, not taylor. Just the fact that $f(x) = f(0) + f'(0) x + o(x)$ where $\lim_{x\to 0} \frac{o(x)}{x} = 0.$ – 5xum Jun 24 '14 at 09:39
  • Isn't $f(x) = f(0) + f'(0) x + o(x)$ a taylor expansion with little o notation ? Isn't $\lim_{x\to 0} \frac{1}{x} = \infty $ ? Lastly, getting rid of the little o is only possible with taking a limit of the whole expression, how does that help here ? – GinKin Jun 24 '14 at 10:08