Integral inequality : for $f\in C^{4}[0,1]$ with $f(0)=f(1)=f^\prime(0)=f^\prime(1)=0$, show that $\int_0^1|\frac{f^{(4)}(x)}{f(x)}|dx\geq 192$

Question

I was interested in the problem stated below:

Problem 0. $f\in C^{4}[0,1]$ with $f(0)=f(1)=f^\prime(0)=f^\prime(1)=0$, show that $\int_0^1|\frac{f^{(4)}(x)}{f(x)}|dx\geq 192$

This problem is assigned as supplementary problems based on the easier one :

Problem 1: $f\in C^{2}[0,1]$ with $f(0)=f(1)=0$, show that $\int_0^1|\frac{f^{\prime\prime}(x)}{f(x)}|dx\geq 4$

I proved the easier one in this way:

Proof of Problem 1: $$\int_{[0,1]}|f''(x)|dx\geq \int_{[\theta_1,\theta_2]}|f''(x)|dx\geq \left|\int_{[\theta_1,\theta_2]}f''(x)dx\right|=|f^\prime(\theta_2)-f^\prime(\theta_1)|$$ As $f$ is continuous on a compact set, we can find $x_0$ such that $f(x_0)>f(x)$ for any $x\in [0,1]$. By the mean value theorem, assign $\theta_1$ and $\theta_2$ above to be: $$\frac{f(x_0)-f(0)}{x_0-0}=f^\prime(\theta_1);\quad \frac{f(x_0)-f(1)}{x_0-1}=f^\prime(\theta_2)$$ Then we have: $$|f^\prime(\theta_2)-f^\prime(\theta_1)|=\left|\frac{f(x_0)}{x_0(1-x_0)}\right|\geq 4|f(x_0)|$$ In sum, we have: $$\int_{[0,1]}\left|\frac{f^{\prime\prime}(x)}{f(x)}\right|\geq \int_{[0,1]}\left|\frac{f^{\prime\prime}(x)}{f(x_0)}\right|\geq 4$$

However for Problem 0, I tried MVT and Taylor expansions on $0$ and $1$ but find no clues to continue. Any reasonable thoughts are welcome! Thank you!

Answer 1

For the Problem 0 here I provide one solution inspired by Hermite interpolation (https://sites.math.rutgers.edu/~falk/math373/lecture7.pdf).

Step 1

Let $$\phi(t)=f(t)-\frac{f(x)}{x^2(x-1)^2}t^2(t-1)^2$$ which is fourth differentiable for $t\neq x$ and $t\in(0,1)$ with $$\phi^{(4)}(t)=f^{(4)}(t)-\frac{12f(x)}{x^2(x-1)^2}$$

One may observe $\phi(0)=\phi(1)=\phi(x)=0$, hence by Rolle's theorem, there exists $\theta_1\in(0,x)$ and $\theta_2\in(x,1)$ such that $$\phi'(0)=\phi'(1)=\phi'(\theta_1)=\phi'(\theta_2)=0$$ Again applying Rolle's theorem one may find $\{\psi_1,\psi_2,\psi_3\}$ and $\{\eta_1,\eta_2\}$ such that $$\phi''(\psi_1)=\phi''(\psi_2)=\phi''(\psi_3)=0$$ and $$\phi'''(\eta_1)=\phi'''(\eta_2)=0$$ Finally, there exists at least one $x_0$ such that $\phi^{(4)}(x_0)=f^{(4)}(x_0)-\frac{12f(x)}{x^2(x-1)^2}=0$ given $x\in [0,1]$.

Step 2:

As $f$ is continuous on a compact set, let $\bar{x}$ to be the global maxima of $f$ on $[0,1]$, then plug in $x=\bar{x}$ to the $\phi(t)$ constructed in Step 1. As there exists $\{\eta_1,\eta_2\}$ such that $\phi'''(\eta_1)=\phi'''(\eta_2)=0$, hence:

$$\int_0^1 |f^{(4)}(x)|dx\geq \int_{\eta_1}^{\eta_2} |f^{(4)}(x)|dx\geq|\int_{\eta_1}^{\eta_2} f^{(4)}(x)dx|=|f^{(3)}(\eta_2)-f^{(3)}(\eta_1)|=\left|\frac{f(\bar{x})}{{\bar{x}}^2(\bar{x}-1)^2}12(\eta_2-\eta_1)\right|\geq \left|\frac{12f(\bar{x})}{{\bar{x}}^2(\bar{x}-1)^2}\right| \geq 192|f(\bar{x})| $$

Then easily to check that : $$\int_0^1 \left|\frac{f^{(4)}(x)}{f(x)}\right|dx\geq \int_0^1 \left|\frac{f^{(4)}(x)}{f(\bar{x})}\right|dx \geq 192$$