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$f:[0,1]\to \mathbb{R}$ is a countinous function.

$$\int_0^1f(x) dx =0 \qquad \mbox{ and } \qquad \int_0^1xf(x) dx =0. $$

If $f \ge 0$ ($f\le0$) were true then $\int_0^1f(x) dx \ge0$ ($\int_0^1f(x) dx \le0$). This is a contradiction, we can conclude that $f$ changes sign. By intermediate value property there exist a point $c$ such that $f(c)=0$. This is the first zero.

By using mean value theorems for integrals I can also show that a zero does exist. I can't show that these zeroes are different from each other.

How to show a second zero exists?

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$\newcommand{\d}[1]{\, \mathrm{d} #1}$ Define: $$ F(x) = \int_0^x f(t) \d{t} $$ Then we observe that $F(0) = F(1) = 0$, and: $$ \int_0^1 F(x) \d{x} = \left[xF(x)\right]_{x=0}^{x=1} - \int_0^1 xf(x)\d{x} = 0 $$ Then, as you mentioned above, we have $c \in (0,1)$ such that $F(c) = 0$. We can now apply Mean Value Theorem, where there exist $c_1 \in (0,c)$ and $c_2 \in (c,1)$ such that $F'(c_1) = F'(c_2) = 0$. Since $F'(x) = f(x)$, $c_1$ and $c_2$ are two distinct zeroes of $f$.

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The given constraints ensure $\int_{0}^{1}(x-k)f(x)\,dx = 0$ for any k.
So, let us assume that $f(x)$ has a unique zero at $x_0\in(0,1)$. Up to replacing $f$ with $-f$, we may as well assume that $f(x)<0$ over $[0,x_0)$ and $f(x)>0$ over $(x_0,1]$. In such a case both $$ \int_{0}^{x_0}(x-x_0)f(x)\,dx\qquad \text{and}\qquad \int_{x_0}^{1}(x-x_0)f(x)\,dx $$ are positive, hence their sum $\int_{0}^{1}(x-x_0)f(x)\,dx$ cannot be zero, contradiction.

It is worth mentioning that this gives a rather powerful criterion for root detection, which is the core of the celebrated result of Conrey, giving the $40\%$ of RH: if for some non-negative weight function $\omega(x)$ all the moments $$ M_k = \int_{0}^{1}x^k f(x)\omega(x)\,dx $$ are (approximately) zero up to $k=n$, then $f(x)$ has at least $n$ zeroes in $[0,1]$.

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Assume that it has 0 zeros, than the integrals can't be zero. So ruled out Assume it has 1 zeros namely k then the function changes sign only once or zero times. If it is zero times, the first condition again doesn't hold. So it must change sign. So by first condition the upper area should be equal to lower area. But by second condition, moment of these areas about the point 0 should be equal in magnitude, but two areas have their centres at different distances as in one area all x are less than k and for the other all greater than k, so equal moment is not possible. Hence it should atleast have 2 zeros as a necessary condition. Sufficiency can be tackled by similar line as well. It is very easy to construct an example for the case of having 2 zeros.

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Let there be only one root , and let that be called $x_1$ , now area of graph from $0$ to $x_1$ , and , from $x_1$ to $1$ must be equal , let it be A . Now since magnitude of integral of $xf(x)$ from $0$ to $x_1$ is less than $x_1 A$ and magnitude of integral of $xf(x)$ from $x_1$ to $1$ is greater than $x_1A$ , therefore integral of $xf(x)$ from $0$ to $1$ won't be equal to 0 ..... a contradiction !!

Therefore the function can not have only 1 root , it must have at least two roots .

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  • $\begingroup$ I don't understand why integral xf(x) from 0 to 1 won't be zero ? $\endgroup$
    – Milan
    Commented Jun 27, 2020 at 10:18
  • $\begingroup$ Because one is one is negative and one is positive but their magnitudes are different, so their sum can't be zero. $\endgroup$
    – ARROW
    Commented Jun 27, 2020 at 11:09

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