Differential Inequality with Zero Initial Value

Detailed Solution

We are given that $f$ is continuous on $[0,1]$, differentiable on $(0,1)$, and $$ f(0)=0,\qquad |f'(x)|\le |f(x)| $$ for every $x\in(0,1)$. We will prove that $f(x)=0$ throughout $[0,1]$. **Step 1: Apply the Mean Value Theorem.** Fix any $x\in(0,1]$. Since $f$ is continuous on $[0,x]$ and differentiable on $(0,x)$, the Mean Value Theorem gives $$ f(x)-f(0)=xf'(c) $$ for some $c\in(0,x)$. Using $f(0)=0$ and the given inequality, $$ |f(x)|=x|f'(c)|\le x|f(c)|. $$ Thus, $$ |f(x)|\le x|f(c)|. $$ This inequality alone does not immediately establish that $f(x)=0$, so we use a standard differential-inequality argument. **Step 2: Define an auxiliary function.** Let $$ g(x)=|f(x)|. $$ Then $g$ is continuous on $[0,1]$ and satisfies $g(0)=0$. For $0\le t\le x\le1$, the Mean Value Theorem applied to $f$ on subintervals, together with $|f'(s)|\le |f(s)|$, leads to the integral inequality $$ |f(x)|\le\int_0^x |f'(t)|\,dt \le\int_0^x |f(t)|\,dt. $$ Consequently, $$ g(x)\le\int_0^x g(t)\,dt. $$ Define $$ G(x)=\int_0^x g(t)\,dt. $$ Since $g$ is continuous, $G$ is differentiable and $$ G'(x)=g(x). $$ Therefore, $$ G'(x)\le G(x). $$ Multiplying by $e^{-x}$ gives $$ e^{-x}G'(x)-e^{-x}G(x)\le0, $$ or $$ \frac{d}{dx}\left(e^{-x}G(x)\right)\le0. $$ Thus $e^{-x}G(x)$ is non-increasing on $[0,1]$. Since $G(0)=0$, $$ e^{-x}G(x)\le0. $$ But $G(x)\ge0$, so $$ G(x)=0 $$ for every $x\in[0,1]$. Since $g$ is continuous and non-negative, the identity $$ \int_0^x g(t)\,dt=0 $$ for every $x\in[0,1]$ implies that $g(x)=0$ throughout the interval. Hence, $$ \boxed{f(x)=0\quad\text{for all }x\in[0,1]}. $$ **Verification of the options** - **A is true:** The differential inequality and initial condition force $f$ to vanish identically. - **B is true as a bound in the sense $|f(x)|\le e^x-1$**, but it is weaker than A. Therefore, it is not the strongest conclusion. - **C is true because $f'(x)=0$ throughout $(0,1)$**, so the derivative vanishes on a dense subset. It is also weaker than A. - **D is false because the function must be identically zero, not merely constant.** Therefore, the strongest valid conclusion is $$ \boxed{\mathrm{A}}. $$