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Let $f:\mathbb{R}\to \mathbb{R}$ be continuous such that $$f(x) = \int_0^xf(s)ds $$ and $f(1) = 0$. Can we prove that $f=0$ without resorting to ODE solving techniques?

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    $\begingroup$ If you have a reason for " without using (whatever)", you should state it clearly in the body of your question. Otherwise, it just reads like "do it blindfolded, with your hands tied behind your back". And this site is about mathematics, not about stunts. $\endgroup$
    – user436658
    Nov 27, 2020 at 12:53
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    $\begingroup$ As is you'll get numerous answers just stating $f'=f\implies f(x)=ae^x$ then $a=0$. So maybe you should ask, can we prove $f=0$ without resorting to ODE solving, for instance using only mean value theorem, or without resorting to differentiation using only inequalities (e.g. $f$ continuous, let suppose $f\neq 0$ then $f(x)\ge M>0$ on $[a,b]$...). $\endgroup$
    – zwim
    Nov 27, 2020 at 13:02

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Note that the differential equation you derived implied that $$ \frac{\partial}{\partial y}\left( f(x+y)f(x-y)\right)=f'(x+y)f(x-y)-f(x+y)f'(x-y) =0.$$ And hence that $$f(x+y)f(x-y) = g(x).$$ We can specify the function $g$ by letting $y = 0$ to get $$f(x)^2 = g(x)$$

and hence

$$f(x+y)f(x-y) = f(x)^2.$$

now sub $x=y$ and note that $f(0) = 0$ to get that

$$f(x)^2 =0$$

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  • $\begingroup$ Where is this coming from ? $\endgroup$
    – user65203
    Nov 27, 2020 at 13:22
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Put $g(x) = |f(x)|$. Note that for $x \ge 0,$ we have

$$g(x) = |f(x)| = \left|\int_0^x f(s) {\rm d}s\right| \le \int_0^x |f(s)| {\rm d}s = \int_0^xg(s) {\rm d}s.$$

(The reason for assuming $x \ge 0$ is to have the integration in the "correct" order and not worry about extra modulus signs.)

Thus, we have $$g(x) \le \int_0^x g(s) {\rm d}s.$$

Using Grönwall's inequality, we directly get the answer. I give a proof anyway.


Putting $G(x) := \displaystyle\int_0^x g(s) {\rm d}s,$ we see that $G'(x) \le G(x)$ or $G'(x) - G(x) \le 0.$ Also, note that $g \ge 0$ and thus, $G(x) \ge 0$ for $x \ge 0$.

Multiplying with $\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right),$ we see that $$(G'(x) - G(x))\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right) \le 0$$ or that $$\left(G(x)\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right)\right)' \le 0.$$

Thus, $$G(x)\exp\left(-\displaystyle\int_0^xG(t){\rm d}t\right)$$ is a decreasing function. However, note that the above is a nonnegative function which is $0$ at $x = 0$. Thus, we must have $G(x) = 0$ for all $x \ge 0$ and thus, $g = G' \equiv 0$ on $[0, \infty)$. A similar argument shows the same for $x < 0$ as well. Thus, we get $g \equiv 0$ or that $f \equiv 0$.

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There is a straightforward way to get your result. You will have $$ f(1)=\int_0^1f(s)ds=0. $$ This identity can be solved taking for example $f(s)=Ce^s$ because the exponential has the property to be equal to its integral. Putting this in the above equation will yield $C=0$ and you are done.

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  • $\begingroup$ This only proves that there is no solution of the form $Ce^s$ with $C\ne0$. $\endgroup$
    – user65203
    Nov 27, 2020 at 13:11
  • $\begingroup$ The other route is to say that this is an integral equation with the only solution $f(s)=0$. $\endgroup$
    – Jon
    Nov 27, 2020 at 13:13
  • $\begingroup$ Uniqueness is ruled out. $\endgroup$
    – user65203
    Nov 27, 2020 at 13:18
  • $\begingroup$ No, in the current OP's question, ODEs are ruled out. $\endgroup$
    – Jon
    Nov 27, 2020 at 15:41

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