Problem: Let $f$ be a function that is continuous on $[0,\infty )$ and differentiable on $(0,\infty ).$ If $f(0)=0$ and $\left\vert f^\prime (x) \right\vert \leq \vert f(x) \vert $ for all $x>0$, then prove that $\left\vert f(x) \right\vert \leq e^x$ for all $x$.
Solution: Note that \begin{align*} \vert f(x) \vert & = \left\vert \int _0^x f^\prime (t)~\mathrm{d} t \right\vert \\ & \leq \int _0^x \left\vert f^\prime (t) \right\vert ~\mathrm{d} t \\ & \leq \int _0^x \vert f(t) \vert \mathrm{d} t. \end{align*} If we assume that \[ F(x) = \int _0^x \vert f(t) \vert \mathrm{d} t, \] then $F(x)$ is antiderivative of $\vert f(x) \vert $. We have \begin{align*} \vert f(t) \vert \leq F(t) & \implies \frac{\vert f(t) \vert }{F(t)} \leq 1 \\ & \implies \int _0^x \frac{\vert f(t) \vert }{F(t)} \mathrm{d} t \leq \int _0^{x} \mathrm{d} t \\ & \implies \ln F(x) \leq x \\ & \implies F(x) \leq e^x. \end{align*} Therefore, we proved that \[ \vert f(x) \vert \leq e^x,~ \forall~x > 0. \]