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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 $t\in(0,1)$. Then if $$\text{sgn}\left(\int_0^1g(x)g''(x)\,dx\right)-\text{sgn}\left(\int_0^1\frac1{g(x)}\,dx\right)=\pm2$$ for some constant $K\in\mathbb{R}$, $$g(x)=Kg'(x)^2.$$ Furthermore, $g''(x)=\dfrac1{2K}$ when $x\in\mathbb{R}-\{0, 1\}$.

Solution

We use the Cauchy-Schwarz inequality for integrals with $a=0$ and $b=1$: $$\left[\int_0^1\frac{g'(x)}{\sqrt{g(x)}}\,dx\right]^2\le\left(\int_0^1g'(x)^2\,dx\right)\left(\int_0^1\frac1{g(x)}\,dx\right)\tag{1}$$

Firstly, consider the LHS. Let $u=g(x)\implies du=g'(x)\,dx$; thus LHS is $$\left[\int_0^1\frac{g'(x)}{\sqrt{g(x)}}\,dx\right]^2=\left[\int_{g(0)}^{g(1)}\frac1{\sqrt{u}}\,du\right]^2=\left(\left[2\sqrt{u}\right]_t^1\right)^2=4(1-\sqrt t)^2=T>0 \tag{2}$$ as $g(0)=t$ and $g(1)=1$. Now use integration by parts for the first integral on the RHS: $$\int_0^1g'(x)^2\,dx=\left[g(x)g'(x)\right]_0^1-\int_0^1g(x)g''(x)\,dx=-\int_0^1g(x)g''(x)\,dx\tag{3}$$ as $g'(0)=g'(1)=0$. Plugging $(2)$ and $(3)$ into $(1)$, we get $$-T\ge\left(\int_0^1g(x)g''(x)\,dx\right)\left(\int_0^1\frac1{g(x)}\,dx\right)\tag{4}$$ with equality holding if, and only if, $$\dfrac{g'(x)}{\sqrt{g(x)}}=k\implies g(x)=\frac1{k^2}g'(x)^2=Kg'(x)^2$$ where $k,K\in\mathbb{R}$ are constants. Differentiating this, we have $g'(x)=2Kg'(x)g''(x)$.

We can divide by $g'(x)\neq0$ when $x\neq0, 1$ so for $x\in\mathbb{R}-\{0,1\}$, $$g''(x)=\frac1{2K}.$$

Now back to $(4)$. $$\int_0^1g(x)g''(x)\,dx>0\implies\int_0^1\frac1{g(x)}\,dx<0$$ and $$\int_0^1g(x)g''(x)\,dx<0\implies\int_0^1\frac1{g(x)}\,dx>0$$ This means that the sign of $\int_0^1g(x)g''(x)\,dx$ is always opposite that of $\int_0^1\frac1{g(x)}\,dx$. The result follows.

TheSimpliFire
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  • Is the problem stated correctly? If $g$ is differentiable and $g(0)=0$ then $1/g(x)$ cannot be (Lebesgue-)integrable on $[0,1].$ – Dap Jan 14 '18 at 16:37
  • Thanks for spotting this! I have changed the problem slightly to avoid division by 0 (hence non-integrability). Please check if there are any other flaws. – TheSimpliFire Jan 14 '18 at 18:46

1 Answers1

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$\def\d{\mathrm{d}} \def\sgn{\mathrm{sgn}}$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$$ \int_0^1 \frac{1}{g(x)} \,\d x > 0. $$ Also,\begin{align*} &\mathrel{\phantom{=}} \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\\ &\mathrel{\phantom{=}} - 12t(1 - t) \int_0^1 x \,\d x + 6t(1 - t)\\ &= \frac{24}{5} (1 - t)^2 - 12(1 - t)^2 + 6(1 - t)^2 - 6t(1 - t) + 6t(1 - t)\\ &= -\frac{6}{5} (1 - t)^2 < 0. \end{align*} Therefore,$$ \sgn\left( \int_0^1 g(x) g''(x) \,\d x \right) - \sgn\left( \int_0^1 \frac{1}{g(x)} \,\d x \right) = -2. $$

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

Ѕᴀᴀᴅ
  • 35,369
  • Thank you for the counterexample! Could you work out where my solution is wrong? – TheSimpliFire Jan 17 '18 at 07:48
  • @TheSimpliFire Actually I don't know why the equality in (4) holds at all as is claimed. – Ѕᴀᴀᴅ Jan 17 '18 at 07:53
  • The 'Mathematical Handbook of Tables and Formulas' states the following: $$\left[\int_0^1f(x)g(x),dx\right]^2\le\left(\int_0^1f(x)^2,dx\right)\left(\int_0^1g(x)^2,dx\right)$$ with equality holding if, and only if $\dfrac{f(x)}{g(x)}$ is a constant. – TheSimpliFire Jan 17 '18 at 07:55
  • But here in (4) it's not an equality, no? – Ѕᴀᴀᴅ Jan 17 '18 at 07:57
  • That's why I said 'with equality holding if, and only if $\dfrac{g'(x)}{\sqrt{g(x)}}=k$...' after the formulation of $(4)$. – TheSimpliFire Jan 17 '18 at 08:00
  • And $\mathbb{sgn}\left( \int_0^1 g(x)g''(x) ,\mathrm{d}x \right) - \mathbb{sgn}\left( \int_0^1 \frac{1}{g(x)} ,\mathrm{d}x \right) = ±2$ doesn't imply this equality. – Ѕᴀᴀᴅ Jan 17 '18 at 08:08
  • Yes it does. It implies that exactly one integral is negative and the other is positive, which is what is inferred in $(4)$. – TheSimpliFire Jan 17 '18 at 14:46
  • Let us continue this discussion in chat. – TheSimpliFire Jan 17 '18 at 15:14
  • Note that $$\mathbb{sgn}\left( \int_0^1 g(x)g''(x) ,\mathrm{d}x \right) - \mathbb{sgn}\left( \int_0^1 \frac{1}{g(x)} ,\mathrm{d}x \right) = ±2$$ is equivalent to $$\left( \int_0^1 g(x)g''(x) ,\mathrm{d}x \right)\left( \int_0^1 \frac{1}{g(x)} ,\mathrm{d}x \right) <0$$ but this does NOT imply $$-T=\left(\int_0^1g(x)g''(x),dx\right)\left(\int_0^1\frac1{g(x)},dx\right)$$ – Veridian Dynamics Jan 19 '18 at 18:28
  • @VeridianDynamics Thanks for the clarification. So if I swap $g(x)=Kg'(x)^2$ and $\text{sgn}\left(\int_0^1g(x)g''(x),dx\right)-\text{sgn}\left(\int_0^1\frac1{g(x)},dx\right)=\pm2$ in my statement then the solution would be correct? – TheSimpliFire Jan 19 '18 at 21:08