Let $f$ is continuous on $[a,b]$. There exists positive K such that $|f(x)| \leq K \int_a^x |f(t)|dt$ then $f(x) = 0$. I was trying to prove the statement above, by trying the smallest number c such that $f(x) = 0$ for any $x<c$ and taking integral near at c. But I get stuck and can somebody give me a hint or idea?
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Let $$F(x)=\int_a^x|f(t)|dt$$
Now, $$e^{-Kx}(F'(x)-KF(x))\le0\qquad\forall x\in[a,b]\\ \implies \Big(e^{-Kx}F(x)\Big)'\le0$$ Therefore, $e^{-Kx}F(x)$ is a non-increasing function taking value $0$ at $x=a$. So, it is non-positive $\forall x\in[a,b]$, and therefore is constant zero function, since both $e^{-Kx}$ and $F$ are non-negative.
Martund
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Here is another way:
Let $M=\max_{x \in [a,b]} \lvert f(x) \rvert$.
If $|f(x)| \leq K \int_a^x |f(t)|dt$, then $|f(x)| \le MK(x-a)$, then $|f(x)| \le M{1 \over 2}K^2(x-a)^2$ and so we see $|f(x)| \le {1 \over n!} MK^n (x-a)^n$ for all $n$.
Hence if we fix any $x \in [a,b]$ we see that $f(x) = 0$.
tristan
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copper.hat
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@Martund: How so? If it is zero on every compact interval $[a,b]$ then it is zero on $[a,\infty)$. Not that it matters, the question was about $[a,b]$. – copper.hat Jun 26 '20 at 05:04
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Ok, I was applying this reasoning with $b\to\infty$ in one go. – Martund Jun 26 '20 at 05:05
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I still don't get your point. The question was about an interval $[a,b]$. – copper.hat Jun 26 '20 at 05:07
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Yeah, I was just wondering whether the reasoning can be extended when $b\to\infty$. – Martund Jun 26 '20 at 05:09
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Might be less provocative to phrase it as a question rather than a declaration. – copper.hat Jun 26 '20 at 05:11
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1Sorry, I meant that it cannot be directly applied when $b=\infty$. – Martund Jun 26 '20 at 05:13