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Let $f:[0,1]\to \Bbb R$ be a differentiable function such that $f(0)=0$ and $|f'(x)| \le k|f(x)|\;\forall x \in [0,1]$,($k>0$), then which of the following is always true?

(A) $f(x)=0 \; \forall \; x \in \Bbb R$
(B) $f(x)=0 \; \forall \; x \in [0,1]$
(C) $f(x) \ne 0 \; \forall \; x \in [0,1]$
(D) $f(1) = k$

This question appeared in a test I gave today (its obviously completed). I would love a hint on how to approach this question, and also some insight on how I should have thought about it from the beginning. Since mean value theorems were on syllabus (Lagrange's mean value theorem, Rolle's theorem) so I suspect their use is required, though I don't see how.

Thank you!

Community
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FreezingFire
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1 Answers1

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Since this is a multiple-choice question, we can settle on (B) just from process of elimination: (A) makes no sense because the domain of the function is $[0,1]$, and the constant function $0$ shows that (C) and (D) are false.

But this is a bit unsatisfying, so let's show that (B) is true: define $$ g(x)=f(x)^2e^{-2kx}$$ for $0\leq x\leq 1$. Then $g(0)=0$, and $$ g^{\prime}(x)=2f(x)f^{\prime}(x)e^{-2k x}-2kf(x)^2e^{-2kx}=2e^{-2kx}(f^{\prime}(x)f(x)-kf(x)^2)$$ $$ \leq 2e^{-2kx}(|f^{\prime}(x)||f(x)|-kf(x)^2)\leq 2e^{-2kx}(k|f(x)|^2-kf(x)^2)=0 $$ Therefore $g$ is non-negative and non-increasing on $[0,1]$, so $0\leq g(x)\leq g(0)=0$ for all $x\in[0,1]$. This implies that $f=0$ on $[0,1]$.

carmichael561
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