$|f'(x)|\le k|f(x)|$ for some $k$ and $f(a) = 0$ imply that $f(x)$ is zero on $[a,b]$. 
Let $f: [a,b] \to \mathbb R$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a) = 0$ and $|f'(x)|\le k|f(x)|$ for some $k$, then $f(x)$ is zero on $[a,b]$. 

I tried proving it using Legrange's Mean Value Theorem but couldn't get it.
$f(x)$ is differentiable on $(a,b)$. $f(x)$ is continuous on $[a,b]$. Let $x$ belong to $[a,b]$ s.t $a < x$. Consider the interval $[a,x]$. By Legrange's Mean Value Theorem, 
$$\frac{f(x)-f(a)}{x-a} = f'(t) ; \ \ a \le t \le x.$$
Since $f(a)=0$, 
$$f(x)=(x-a)f'(t)$$
$$|f(x)| \le (x-a)k|f(t)|.$$
After this, I was thinking of proceeding with the inequality by putting $f(a)$ in the place of $f(t)$.
 A: As I commented, we first prove the conclusion for a smaller interval. 
Proof. $\blacktriangleleft$ We may assume $b-a  > 1/(2k)$ [otherwise we proceed on $[a,b]$, and the following argument is still applicable with slight modification] First consider the interval $[a, a + 1/(2k)] \subset [a,b]$. By Lagrange’s MVT, 
$$
\forall x \in \left[a, a + \frac {b-a}{2k}\right],\;|f(x)| = |f(x) - f(a)| = |f’(c_1)||x-a| \leqslant k |f(c_1)| \cdot \frac {1}{2k} = \frac 12 |f(c_1)| \quad [c_1\in (a, x)].
$$
Now for $c_1$, the Lagrange’s MVT gives that
$$
|f(c_1)| = |f(c_1) - f(a)| = |f’(c_2)| |c_1 - a| \leqslant k |f(c_2)| \cdot \frac {1} {2k} = \frac 12 |f(c_2)| \quad [c_2 \in (a,c_1)].
$$
Repeatedly we have a sequence $(c_n)_1^\infty$ s.t. $0 < c_n < c_{n-1} < x$ and 
$$
|f(c_{n-1})| \leqslant \frac 12 |f(c_{n})|\quad [\forall n \in \mathbb N^*],
$$
hence
$$
\forall n, |f(x)| \leqslant \frac 12 |f(c_1)| \leqslant \frac 14 |f(c_2)| \leqslant \cdots \leqslant \frac 1{2^n} |f(c_n)|.
$$
Since $f$ is continuous on $[a, b]$, $|f|$ is also continuous and attain its maximum on $[a, b]$. Let the maximum be $M$, then 
$$
\forall n \in \mathbb N^*, |f(x)| \leqslant \frac M {2^n}.
$$
Let $n \to \infty$, then $|f| \equiv 0$ on $[a, a+1/(2k)]$. 
Now for the whole interval $[a,b]$, divide it into $m = \lceil (b-a)/2k \rceil$ subintervals $[a + (j-1)/(2k), a+j/(2k)] [j \leqslant m-1]$ and $[a + (m-1)/2k, b]$, then by induction we could successively show that $f$ vanishes on each subinterval [because $f$ vanishes at the left endpoint of each subinterval]. Hence $f \equiv 0$ on $[a,b]$. $\blacktriangleright$
A: Since $f$ is continuous on $[a,b]$ and $f(a)=0,$ if $f$ is not $0$ on $[a,b]$ then $f(c)\ne 0$ for some $c\in (a,b).$
Suppose by contradiction that $c\in (a,b)$ and $ f(c)\ne 0$. Because $f$ is continuous and $f(a)=0$ there exists $d\in [a,c)$  such that $f(d)=0$ and $\forall x\in (d,c]\;(f(x)\ne 0).$
For $x\in (d,c] $ let $g(x)=\ln |f(x)|.$ Since $f$ does not change sign on $(d,c]$ we have $g'(x)=f'(x)/f(x)$ for all $x\in (d,c]$. Hence $|g'(x)|\leq k$ for all $x\in (d,c].$
Let $(x_n)_{n\in \Bbb N }$ be a strictly decreasing sequence converging to $d,$ with $x_1=c.$ For each $n\in \Bbb N$ there exists $y_n\in (x_{n+1},x_1)$ such that  $$\left|\frac {g(x_1)-g(x_{n+1})}{x_1-x_{n+1}}\right|=|g'(y_n)|\leq k.$$ So for all $n\in \Bbb N$ we have $$(\bullet) \quad |g(x_1)-g(x_{n+1})|\leq k|x_1-x_{n+1}|=k|c-x_{n+1}|=c-x_{n+1}<c-d.$$ But $f$ is continuous and $x_{n+1}\to d$ as $n\to \infty,$ so $|f(x_{n+1})|\to |f(d)|=0$ as $n\to \infty.$ This implies that $|g(x_{n+1})|=|\;\ln |f(x_{n+1})|\;|\to \infty$ as $n\to \infty,$ contradictory to  $(\bullet ).$
A: Since $f\left(a\right)=0$ and $f $ is continuous then exists an interval $\left[a,a_{1}\right]$ where $f$ does not change sign. Now assume that $f\left(x\right)\geq0$ in this interval. Then, since $$\left|f^{\prime}\left(x\right)\right|\leq kf\left(x\right)\Rightarrow f^{\prime}\left(x\right)\leq kf\left(x\right)$$ we have, by the Gronwall's inequality, $$f\left(x\right)\leq f\left(a\right)\exp\left(k(x-a)\right),\,\forall x\in\left[a,a_{1}\right]$$ and so $f\equiv0$. If $f\left(x\right)\leq0$ we note that $$\left|f^{\prime}\left(x\right)\right|\leq-kf\left(x\right)\Rightarrow-f^{\prime}\left(x\right)\leq-kf\left(x\right)$$ then, again by the Gronwall's inequality applied to $-f$, we get $$-f\left(x\right)\leq-f\left(a\right)\exp\left(k(x-a)\right)$$ and so $f\equiv0$ (remember that we assumed that $f$ is negative). If $f$ does not change sign in $\left[a,b\right]$ we have done; if $f$ changes sign by the intermediate value theorem exist some $a_{2}\in\left[a,b\right]$ such that $f\left(a_{2}\right)=0$. Then we apply the technique in each sub-interval.
A: Seeing that there was some confusing about my suggestion in the comments, I'll explain more fully:
We have by assumption
$$
|\frac{d}{dx}\ln f(x)| \le k
$$
wherever this is well defined. Since $f(a) = 0$ and $|\ln 0| = \infty$, we also know $\ln f(x)$ must be unbounded on $(a,b)$. 
Suppose $\exists c\in [a,b]$ such that $f(c) \ne 0$. Without loss of generality, assume $f(c) > 0$ (if not, do the following with $-f(x)$ instead). 
$$
|\ln f(x) - \ln f(c)| = |\int_c^x \frac{d}{dt}\ln f(t) dt| \le \int_c^x |\frac{d}{dt}\ln f(t)| dt \le k |x-c| \le k (b-a)
$$
This contradicts unboundedness of $\ln f(x)$, so we must conclude $f(x) = 0$ for all $x \in [a,b]$.
