Formally show that $f(0)=0$, $|f'(x)|\leq|f(x)|$ on $[0,1]$ implies $f(x)=0$ on $[0,1]$ Assume continuity on $[0,1]$ and differentiability on $(0,1)$. So far I have reasoned as follows: 
If $f(c)\neq 0$ for some $c$, then $f'(a)\neq 0$ for some $a$. WLOG assuming positivity, $f'(a)a<f(a)$ (as $a<1$ and $f'(a)\leq f(a))$. This means that $f'(a)<f(a)/a=(f(a)-0))/(a-0)$. This means that the slope at every point is "flatter" than the secant line between that point and $(0,0)$. I feel like this would prevent a limit toward $0$ and thus be in contradiction with the continuity condition. 
However, I am getting stuck trying to rigorously prove this. Is another proof needed altogether? 
 A: The mean value theorem is what you're after.
Assume $f(x)$ is not $0$ on the whole interval, and take some $a_0\in (0,1)$ such that $f(a_0)\neq 0$. Then by the mean value theorem there must be some $a_1\in (0,a_0)$ such that
$$
f'(a_1) = \frac{f(a_0) - f(0)}{a_0-0} = \frac{|f(a_0)|}{a_0}
$$
By the assumed property of $f$, we have $|f(a_1)|\geq \frac{|f(a_0)|}{a_0}$.
Now again by the mean value theorem, there is an $a_2\in (0, a_1)$ such that
$$
f'(a_2) = \frac{f(a_1) - f(0)}{a_1-0}
$$
and again we get $|f(a_2)|\geq\frac{|f(a_1)|}{a_1}\geq \frac{|f(a_0)|}{a_0^2}$.
And we continue this pattern. Each step we create a new $a_n$ for which $|f(a_n)|\geq \frac{|f(a_0)|}{a_0^n}$. Since $a_0<1$, this means that $|f(a_n)|\to \infty$. But a continuous function on a closed and bounded interval must be bounded. So we reach a contradiction.
A: \begin{align*}
|f(x)|&=\left|\int_{0}^{x}f'(t_{1})dt_{1}\right|\\
&\leq\int_{0}^{x}|f(t_{1})|dt_{1}\\
&\leq\int_{0}^{x}\int_{0}^{t_{1}}|f(t_{2})|dt_{2}dt_{1}\\
&\leq\int_{0}^{x}\int_{0}^{t_{1}}\int_{0}^{t_{2}}|f(t_{3})|dt_{3}dt_{2}dt_{1}\\
&\leq\cdots\\
&\leq\int_{0}^{x}\int_{0}^{t_{1}}\cdots\int_{0}^{t_{n-1}}|f(t_{n})|dt_{n}\cdots dt_{2}dt_{1}\\
&\leq\left(\max_{[0,x]}|f|\right)\int_{0}^{x}\int_{0}^{t_{1}}\cdots\int_{0}^{t_{n-1}}dt_{n}\cdots dt_{2}dt_{1}\\
&=\left(\max_{[0,x]}|f|\right)\dfrac{1}{n!}x^{n},
\end{align*}
now taking $n\rightarrow\infty$.
A: There’s already an accepted answer, but I wanted to provide a direct proof not relying on any iterative construction. 
Assume first that $f$ takes both positive and negative values with $f(1) \neq 0$. Then, depending on the sign of $f(1)$, the function $h(x)=xf(x)$ has a global extremum at a point $0 < c < 1$ (take the extremum of sign different from $f(1)$), thus $h(c) \neq 0$ hence $f(c)=0$.  Therefore $0=h’(c)=cf’(c)+f(c)$, thus $|f’(c)| > |f(c)|$, a contradiction. 
Similarly, if $f \neq 0$ with $f(1)=0$, $h(x)=xf(x)$ has a global extremum $0 \leq c \leq 1$ with $g(c) \neq 0$, so $f(c) \neq 0$ and we get a contradiction by computing $h’(c)$ just as above.
So we may assume $f \geq 0$. 
Let $g(x)=\frac{f(x)}{x}$ for $0 < x \leq 1$. It is easy to show that for $0 < x < 1$ $g’(x) \leq 0$, so $g$ nonincreasing. By the MVT and the assumption, $g$ goes to $0$ at $0$, thus $g \leq 0$. Since $f \geq 0$, $f=0$. 
