Let $f\in C^1[a,b]$, $f(a)=0$ and suppose that $\lambda\in\mathbb R$, $\lambda >0$, is such that
$$|f'(x)|\leq \lambda |f(x)|$$
for all $x\in [a,b]$. Is it true that $f(x)=0$ for all $x\in [a,b]$?
1. **Define the set \( A \):**
   Let \( A = \{ x \in [a,b] \mid f(x) = 0 \} \). Since \( f(a) = 0 \), we have \( a \in A \), so \( A \) is non-empty.

2. **Show that \( A \) is open:**
   Suppose \( x \in A \). Then \( f(x) = 0 \). We need to show that there exists an interval around \( x \) where \( f \) is also zero.

   Since \( f \in C^1[a,b] \), \( f \) is continuously differentiable, and we have the condition \( |f'(x)| \leq \lambda |f(x)| \) for all \( x \in [a,b] \).

3. **Consider a small interval around \( x \):**
   Let \( \delta > 0 \) be small enough such that \( [x - \delta, x + \delta] \subset [a,b] \). For any \( y \in [x - \delta, x + \delta] \), we can use the given inequality \( |f'(y)| \leq \lambda |f(y)| \).

4. **Use the inequality to bound \( f(y) \):**
   Since \( f(x) = 0 \), we can consider the maximum value of \( |f(y)| \) in the interval \( [x - \delta, x + \delta] \). Let \( M = \max_{y \in [x - \delta, x + \delta]} |f(y)| \).

   By the given condition, for any \( y \in [x - \delta, x + \delta] \),
   \[
   |f'(y)| \leq \lambda |f(y)|
   \]

5. **Iterate the inequality:**
   Repeating the inequality, we get
   \[
   |f(y)| \leq (2\delta \lambda)^n M
   \]
   for some \( n \). If \( \delta \) is chosen such that \( 2\delta \lambda < 1 \), then as \( n \to \infty \), \( (2\delta \lambda)^n \to 0 \). Hence, \( |f(y)| \to 0 \).

6. **Conclude that \( f(y) = 0 \):**
   Therefore, \( f(y) = 0 \) for all \( y \in [x - \delta, x + \delta] \). This shows that \( A \) is open.

7. **Show that \( A \) is closed:**
   Since \( A \) is open and \( [a,b] \) is a closed interval, the complement of \( A \) in \( [a,b] \) is also open. But \( A \) is non-empty and \( f \) is continuous, so \( A \) must be both open and closed in the connected interval \( [a,b] \).

8. **Conclude that \( A = [a,b] \):**
   Since \( [a,b] \) is connected and \( A \) is both open and closed, it follows that \( A = [a,b] \).

Therefore, \( f(x) = 0 \) for all \( x \in [a,b] \).

\(\blacksquare\)

The final answer is \( \boxed{ f(x) = 0 } \) for all \( x \in [a,b] \).