Vtyjkyuk

I discovered this yesterday and I just want to know whether my solution is correct and whether there's a shortcut to it.

Let $g(x)$ be a twice-differentiable continuous function that crosses the points $(0, t)$ and $(1, 1)$ and has stationary points there with $tin(0,1)$. Then if $$textsgnleft(int_0^1g(x)g''(x),dxright)-textsgnleft(int_0^1frac1g(x),dxright)=pm2$$ for some constant $KinmathbbR$, $$g(x)=Kg'(x)^2.$$ Furthermore, $g''(x)=dfrac12K$ when $xinmathbbR-0, 1$.

Solution

We use the Cauchy-Schwarz inequality for integrals with $a=0$ and $b=1$:
$$left[int_0^1fracg'(x)sqrtg(x),dxright]^2leleft(int_0^1g'(x)^2,dxright)left(int_0^1frac1g(x),dxright)tag1$$

Firstly, consider the LHS. Let $u=g(x)implies du=g'(x),dx$; thus LHS is $$left[int_0^1fracg'(x)sqrtg(x),dxright]^2=left[int_g(0)^g(1)frac1sqrtu,duright]^2=left(left[2sqrturight]_t^1right)^2=4(1-sqrt t)^2=T>0 tag2$$ as $g(0)=t$ and $g(1)=1$. Now use integration by parts for the first integral on the RHS: $$int_0^1g'(x)^2,dx=left[g(x)g'(x)right]_0^1-int_0^1g(x)g''(x),dx=-int_0^1g(x)g''(x),dxtag3$$ as $g'(0)=g'(1)=0$. Plugging $(2)$ and $(3)$ into $(1)$, we get $$-Tgeleft(int_0^1g(x)g''(x),dxright)left(int_0^1frac1g(x),dxright)tag4$$ with equality holding if, and only if, $$dfracg'(x)sqrtg(x)=kimplies g(x)=frac1k^2g'(x)^2=Kg'(x)^2$$ where $k,KinmathbbR$ are constants. Differentiating this, we have $g'(x)=2Kg'(x)g''(x)$.

We can divide by $g'(x)neq0$ when $xneq0, 1$ so for $xinmathbbR-0,1$, $$g''(x)=frac12K.$$

Now back to $(4)$. $$int_0^1g(x)g''(x),dx>0impliesint_0^1frac1g(x),dx<0$$ and $$int_0^1g(x)g''(x),dx<0impliesint_0^1frac1g(x),dx>0$$ This means that the sign of $int_0^1g(x)g''(x),dx$ is always opposite that of $int_0^1frac1g(x),dx$. The result follows.

$defdmathrmd defsgnmathrmsgn$For a fixed $t in (0, 1)$, take$$
g(x) = -2(1 - t) x^3 + 3(1 - t) x^2 + t,
$$
so $g(0) = t$ and $g(1) = 1$. Because$$
g'(x) = 6(1 - t) x(1 - x),
$$
then $x = 0$ and $x = 1$ are stationary points of $g(x)$ and $g$ is increasing on $[0, 1]$. Thus for $x in [0, 1]$, $g(x) geqslant g(0) = t > 0$, so$$
int_0^1 frac1g(x) ,d x > 0.
$$
Also,beginalign*
&mathrelphantom= int_0^1 g(x) g''(x) ,d x\
&= 6(1 - t) int_0^1 (-2(1 - t) x^3 + 3(1 - t) x^2 + t)(-2x + 1) ,d x\
&= 24(1 - t)^2 int_0^1 x^4 ,d x - 48(1 - t)^2 int_0^1 x^3 ,d x + 18(1 - t)^2 int_0^1 x^2 ,d x\
&mathrelphantom= - 12t(1 - t) int_0^1 x ,d x + 6t(1 - t)\
&= frac245 (1 - t)^2 - 12(1 - t)^2 + 6(1 - t)^2 - 6t(1 - t) + 6t(1 - t)\
&= -frac65 (1 - t)^2 < 0.
endalign*
Therefore,$$
sgnleft( int_0^1 g(x) g''(x) ,d x right) - sgnleft( int_0^1 frac1g(x) ,d x right) = -2.
$$

However, because $deg g = 3$ and $deg (g')^2 = 2 deg g' = 4$, there does not exist a constant $K$ such that$$
g(x) = K(g'(x))^2. quad forall x in (0, 1)
$$