Romanian Mathematical Olympiad

$${\int }_{-1}^1{x}^{2n}f(x)dx={\int }_{-1}^1{x}^{2n}g(x)dx+{\int }_{-1}^1{x}^{2n}h(x)dx=2{\int }_0^1{x}^{2n}g(x)dx.$$ Since $f(0)=0$ and $g^{\prime }(x)=\frac{1}{2}(f^{\prime }(x)-f^{\prime }(-x))$, it follows that $g(0)=g^{\prime }(0)=0$; and since $g^{\prime \prime }(x)=\frac{1}{2}(f^{\prime \prime }(x)+f^{\prime \prime }(-x))$ and $f^{\prime \prime }$ is continuous, so is $g^{\prime \prime }$. Let $m={\min }_{0\le x\le 1}{g}^{\prime \prime }(x)$ and let $M={\max }_{0\le x\le 1}{g}^{\prime \prime }(x)$. By Lagrange's theorem, for each $t$ in $(0,1]$, there exists a $c_t$ in $(0,t)$ such that $g^{\prime }(t)=t{g}^{\prime \prime }(c_t)$, so $mt\le g^{\prime }(t)\le Mt$. Integrate the latter over the interval $[0,x]$, where $0<x\le 1$, to get $\frac{1}{2}m{x}^2\le g(x)\le \frac{1}{2}M{x}^2$, so $\frac{1}{2}m{x}^{2n+2}\le x^{2n}g(x)\le \frac{1}{2}M{x}^{2n+2}$ for all $x$ in $[0,1]$. Integration of the latter over $[0,1]$ yields $$\frac{m}{2(2n+3)}={\int }_0^1\frac{1}{2}m{x}^{2n+2}dx\le {\int }_0^1{x}^{2n}g(x)dx\le {\int }_0^1\frac{1}{2}M{x}^{2n+2}dx=\frac{M}{2(2n+3)},$$ $$\text{so}$$ $$m\le 2(2n+3){\int }_0^1{x}^{2n}g(x)dx\le M,\text{i. e.,}m\le (2n+3){\int }_{-1}^1{x}^{2n}f(x)dx\le M.$$

Since $g^{\prime \prime }$ has the intermediate value property (Darboux), $$g^{\prime \prime }(b)=(2n+3){\int }_{-1}^1{x}^{2n}f(x)dx,\text{for some }b\text{ in }[0,1].$$ Finally, since $g^{\prime \prime }(b)$ lies between $f^{\prime \prime }(-b)$ and $f^{\prime \prime }(b)$, and $f^{\prime \prime }$ has the intermediate value property (Darboux), $$f^{\prime \prime }(c)=g^{\prime \prime }(b)=(2n+3){\int }_{-1}^1{x}^{2n}f(x)dx,\text{for some }c\text{ in }[-b,b]\subseteq [-1,1].$$

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Alternative solution.

Alternative Solution. Dropping the condition that $f^{\prime \prime }$ be continuous, we establish the required equality for some $c$ in the open interval $(-1,1)$. Letting $m={\inf }_{|x|\le 1}{f}^{\prime \prime }(x)$ and $M={\sup }_{|x|\le 1}{f}^{\prime \prime }(x)$, we show that $$m\le (2n+3){\int }_{-1}^1{x}^{2n}f(x)dx\le M.(\ast )$$ Equality holds at one end if and only if it holds at the other, in which case $f$ is a degree 2 polynomial function vanishing at the origin; otherwise, both inequalities are strict. Since $f^{\prime \prime }$ has the intermediate value property (Darboux), the conclusion follows: The case where $f$ is a degree 2 polynomial function vanishing at the origin is clear — any point in the open interval $(-1,1)$ will do. Otherwise, choose $\alpha$ and $\beta$ such that $$m<\alpha <(2n+3){\int }_{-1}^1{x}^{2n}f(x)dx<\beta <M.$$ Then $\alpha =f^{\prime \prime }(a)$ and $\beta =f^{\prime \prime }(b)$ for some $a$ and $b$ in $[-1,1]$. Clearly, $a$ and $b$ are distinct, so $(2n+3){\int }_{-1}^1{x}^{2n}f(x)dx=f^{\prime \prime }(c)$, for some $c$ strictly between $a$ and $b$; that is, $c$ lies in the open interval $(-1,1)$. To prove $(\ast )$, consider the 3-term power expansion of $f$ at the origin: $$f(x)=f(0)+f^{\prime }(0)\cdot x+\frac{1}{2}{f}^{\prime \prime }(t)\cdot x^2=f^{\prime }(0)\cdot x+\frac{1}{2}{f}^{\prime \prime }(t)\cdot x^2,$$ for some $t$ between 0 and $x$. Thus, $\frac{1}{2}m{x}^2\le f(x)-f^{\prime }(0)\cdot x\le \frac{1}{2}M{x}^2$ for all $x$ in $[-1,1]$. Multiplication by $(2n+3)x^{2n}\ge 0$ and integration over $[-1,1]$ yields $(\ast )$. To complete the argument, assume equality at the left end; the other case is dealt with similarly. Then $m$ is finite, and $x\mapsto f(x)-f^{\prime }(0)\cdot x-\frac{1}{2}m{x}^2$, $|x|\le 1$, is a non-negative continuous function whose integral over $[-1,1]$ vanishes. Consequently, this function vanishes identically, so $f$ is a degree 2 polynomial function vanishing at the origin, and equality holds at the other end as well. This completes the argument and concludes the proof.