Math Genius: Prove that $exists a, bin(0,1)$ such that $int_0^{a} xf(x)dx=0text{ and }int_0^bxf(x)dx=frac{b^2f(b)}{2}.$

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Question: Let $f:[0,1]tomathbb{R}$ be a continuous function such that $f(0)=0$ and $$int_0^1f(x)dx=0.$$ Prove that $exists a, bin(0,1)$ such that

$$int_0^{a} xf(x)dx=0text{ and }int_0^bxf(x)dx=frac{b^2f(b)}{2}.$$

My approach: Let $g:[0,1]tomathbb{R}$ be such that $$g(x)=xint_0^xf(t)dt-int_0^xtf(t)dt, forall xin[0,1].$$

By the first fundamental theorem of calculus we can conclude that $g$ is differentiable on $[0,1]$ and $$g'(x)=int_0^xf(t)dt, forall xin[0,1].$$

Also, observe that $g(0)=0$ and $g(1)=-int_0^1tf(t)dt$. Thus, by applying MVT to the function $g$ on the interval $[0,1]$, we can conclude that, $exists cin(0,1)$ such that $$g'(c)=int_0^cf(t)dt=-int_0^1tf(t)dt.$$

Observe that clearly three cases are possible, i.e, either$$int_0^cf(t)dt<0text{ or }int_0^cf(t)dt=0text{ or }int_0^cf(t)dt>0.$$

Now let $h:[0,1]tomathbb{R}$ be such that $$h(x)=int_0^xtf(t)dt, forall xin[0,1].$$

Please note that the part highlighted below is wrong, but still I have included it to just demonstrate my thinking process, as it might be of some help to others trying out this problem.

Observe that if $int_0^cf(t)dt<0$, then, $h(1)>0$. This also implies that $exists$ an open interval $(d,e)in[0,c]$, such that $f(t)<0, forall tin(d,e)$. Now select any point $c_1in(d,e)$. Applying MVT to the function $h$ on the interval $[0,c_1]$, we can conclude that $exists c_2in(0,c_1)$ such that $$h'(c_2).c_1=f(c_2).c_2.c_1=h(c_1)-h(0)=h(c_1)<0.$$

Now $h(c_1)<0$ and $h(1)>0$. Thus, by IVT we can conclude that $exists ain(c_1,1)subseteq(0,1)$, such that $$h(a)=int_0^af(t)dt=0.$$

A similar reasoning for the case when $int_0^cf(t)dt>0,$ shows that $exists ain(0,1)$, such that $$h(a)=int_0^af(t)dt=0.$$

Now finally if $int_0^cf(t)dt=0$, then we will have $h(1)=0$. Now if $f$ is identically equal to $0$ on $[0,c]$, then clearly $tf(t)=0, forall tin[0,c]implies h(x)=0, forall xin[0,c].$ Thus choosing any point $xin(0,c]$ and setting it as $a$, we will have $h(a)=0$ and we will be done in that case.

Now if $f$ acquires both positive and negative values on $[0,c]$, then we can conclude that $exists c_1,c_2in(0,c)$, such that $f(c_1)>0$ and $f(c_2)<0$. Also, let us assume WLOG that $c_2>c_1$.

I have not been able to make any significant approach other than this. Can someone help me out with this problem? Please note that a solution using integration by parts might not be possible, since $f$ is not a differentiable function.

The first part is answered in Olympiad calculus problem. We can build on Christian Blatter’s answer to solve the second part.

Define $F(x) = int_0^x f(t) , dt$ and $phi(x) = frac 1x int_0^x F(t), dt$, $phi(0) = 0$. In Christian’s answer it is demonstrated that
$$
phi'(x) = frac{1}{x^2} int_0^x tf(t) , dt
$$

has a zero $a in (0, 1)$, so that $int_0^a tf(t) , dt = 0$.

In our case we additionally have $f(0) = 0$, which implies that $phi$ is differentiable at $x=0$ with $phi'(0) = 0$. Applying the mean-value theorem to $phi’$ gives that
$$
phi”(x) = -frac{2}{x^3} int_0^x tf(t) , dt + frac{f(x)}{x}
$$

has a zero $b in (0, a)$, so that $int_0^b tf(t) , dt = frac{b^2f(b)}{2}$.

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