Differential Inequality and Zero Function

Detailed Solution

We are given $$ f(0)=0 $$ and $$ |f'(x)|\le |f(x)| $$ for every $x\in\mathbb R$. We will prove that $f(x)=0$ for every real number $x$. **Step 1: Show that $f'(0)=0$.** Putting $x=0$ into the inequality gives $$ |f'(0)|\le |f(0)|=0. $$ Therefore, $$ f'(0)=0. $$ However, this alone does not prove that $f$ is identically zero. We need to use the differential inequality on the entire real line. **Step 2: Consider an arbitrary $x>0$.** By the Fundamental Theorem of Calculus, since $f$ is differentiable and $|f'(t)|\le |f(t)|$, we need to be careful: differentiability alone does not guarantee that $f'$ is continuous. Instead, apply the Mean Value Theorem to obtain, for every $t\in[0,x]$, $$ |f(t)|\le \int_0^t |f'(s)|\,ds $$ provided the integral argument is justified. A more direct approach uses a standard differential inequality result. Since $f$ is differentiable, it is continuous. Define $$ g(x)=f(x)^2. $$ Then $g$ is differentiable, and $$ g'(x)=2f(x)f'(x). $$ Using the given inequality, $$ |g'(x)| =2|f(x)||f'(x)| \le 2f(x)^2 =2g(x). $$ Also, $g(x)\ge0$ and $g(0)=0$. For $x\ge0$, the inequality $g'(x)\le2g(x)$ implies that $$ \frac{d}{dx}\left(e^{-2x}g(x)\right) =e^{-2x}\bigl(g'(x)-2g(x)\bigr)\le0. $$ Thus, $e^{-2x}g(x)$ is non-increasing on $[0,\infty)$. Since $g(0)=0$ and $g(x)\ge0$, we obtain $$ 0\le e^{-2x}g(x)\le g(0)=0. $$ Hence $g(x)=0$ for every $x\ge0$, so $$ f(x)=0\qquad(x\ge0). $$ **Step 3: Consider $x<0$.** Define $$ h(t)=f(-t). $$ Then $h(0)=0$ and $$ |h'(t)|=|f'(-t)|\le |f(-t)|=|h(t)|. $$ Applying the preceding argument to $h$ for $t\ge0$ gives $h(t)=0$. Therefore, $$ f(x)=0\qquad(x\le0). $$ Combining both cases, $$ \boxed{f(x)=0\quad\text{for every }x\in\mathbb R}. $$ **Verification of the options** - A is false because $f$ must vanish on the entire real line, not merely on $[0,1]$. - B is true because the differential inequality and the initial condition force $f$ to be identically zero. - C is false because no nonzero function can satisfy the given conditions. - D is false because, although $f'(0)=0$, the function must also be identically zero. Therefore, the correct option is $$ \boxed{\mathrm{B}}. $$