Prove integral inequality including n-order derivative

Question

Let $f$ be $n$ times continuously differentiable on $[0,1]$, with $f(\frac{1}{2})=0$ and $f^{(i)}(\frac{1}{2})=0$ when $i$ is even and less than $n$. Prove $$ \left( \int_0^1{f\left( x \right) \mathrm{d}x} \right) ^2\leqslant \frac{1}{\left( 2n+1 \right) 4^n\left( n! \right) ^2}\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}. $$

$f\left( x \right) =f\left( \frac{1}{2} \right) +f'\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) +\frac{1}{2!}f''\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) ^2+\cdots +\frac{1}{n!}f^{\left( n-1 \right)}\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) ^{n-1}+\frac{1}{n!}f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n=\frac{1}{n!}f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n$，$\theta\in(0,1)$.

so $\int_0^1{f\left( x \right) \mathrm{d}x}=\frac{1}{n!}\int_0^1{f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n\mathrm{d}x}$

How to deal with $\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}$ to prove this inequality?

Answer 1

Your solution does not work because $\theta$ (from the mean-value form of the remainder) depends on $x$.

As @leoli1 already said, one has to use Taylor's theorem with the integral remainder, this gives $$ \int_0^1f(x) \, dx=\int_0^1\int_{1/2}^x \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t)\, dt \, dx \, . $$ Now we split the integral in two parts so that we can interchange the order of integration. First, $$ \int_{1/2}^1\int_{1/2}^x \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t)\, dt \, dx \\ = \int_{1/2}^1\int_t^1 \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dx\, dt = \frac{1}{n!} \int_{1/2}^1 (1-t)^n f^{(n)}(t)\, dt \, . $$ Second, $$ \int_0^{1/2}\int_{1/2}^x \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t)\, dt \, dx \\ = -\int_0^{1/2}\int_0^t \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dx\, dt = \frac{(-1)^n}{n!} \int_0^{1/2} t^nf^{(n)}(t) \, dt \, . $$ Combining these equations we get $$ \int_0^1f(x) \, dx = \frac{1}{n!} \left( (-1)^n \int_0^{1/2} t^nf^{(n)}(t) \, dt + \int_{1/2}^1 (1-t)^n f^{(n)}(t)\, dt\right) \\ \le \frac{1}{n!} \int_0^1 \min(t, 1-t)^n |f^{(n)}(t)| \, dt \, . $$ Now we can apply the Cauchy-Schwarz inequality: $$ \left( \int_0^1f(x) \, dx \right) \le \frac{1}{n!^2} \int_0^1 \min(t, 1-t)^{2n} \, dt \int_0^1 |f^{(n)}(t)|^2 \, dt \, . $$ This gives the desired estimate because $$ \int_0^1 \min(t, 1-t)^{2n} \, dt = 2 \int_0^{1/2} t^{2n} \, dt = \frac{1}{(2n+1)4^n} \, . $$

Answer 2

This almost achieves the desired bound, maybe I have some calculations error in here and it actually works.

For $n=0$ this is just Cauchy-Schwarz, so assume $n>0$.

Let $a=\frac{1}{2}$ for clarity and $x\in[0,1]$. Define $$R_{n-1}(x)=\int_a^x\frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}dt$$ This is the error term in Taylor's theorem, i.e. $$f(x) = f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1}+R_{n-1}(x)$$ Therefore:$$\int_0^1f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}\int_0^1(x-a)^{k}dx+\int_0^1R_{n-1}(x)dx$$ Note that every term in the sum vanishes: For even $k$ because of $f^{(k)}(a)=0$ and for odd $k$ because the integral is zero due to symmetry.
Thus $$\int_0^1f(x)dx=\int_0^1R_{n-1}(x)dx=\int_0^1\int_a^x\frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}dtdx$$ Now by Cauchy-Schwarz we have \begin{align*} \left\vert\int_a^x{f^{(n)}(t)}(x-t)^{n-1}dt\right\vert^2\leq\left\vert\int_a^x{f^{(n)}(t)}^2dt\right\vert\cdot\left\vert\int_a^x(x-t)^{2(n-1)}dt\right\vert \end{align*} The absolute values are necessary on the right side to include the case $x<a$. Now we have$$\left\vert\int_a^x{f^{(n)}(t)}^2dt\right\vert\leq\int_0^1f^{(n)}(t)^2dt$$ and \begin{align*} \left\vert\int_a^x(x-t)^{2(n-1)}dt\right\vert=\left\vert\frac{(x-a)^{2n-1}}{2n-1}\right\vert \end{align*} Hence \begin{align*} (n-1)!\int_0^1f(x)dx&\leq\int_0^1\left\vert\int_a^xf^{(n)}(x-t)^{n-1}dt\right\vert dx\\ &\leq\int_0^1\sqrt{\int_0^1f^{(n)}(t)^2dt}\sqrt{\left\vert\frac{(x-a)^{2n-1}}{2n-1}\right\vert}dx \end{align*} And: \begin{align*} \int_0^1\sqrt{\left\vert\frac{(x-a)^{2n-1}}{2n-1}\right\vert}dx&=\frac{2}{\sqrt{2n-1}}\int_0^{a}x^{(2n-1)/2}dx\\ &=\frac{2}{\sqrt{2n-1}}\frac{\left(\frac{1}{2}\right)^{(2n+1)/2}}{\frac{2n+1}{2}}\\ &=\frac{4}{\sqrt{2n-1}(2n+1)}\frac{1}{2^{(2n+1)/2}} \end{align*} Putting all together then gives us: \begin{align*} \left(\int_0^1f(x)dx\right)^2&\leq \frac{1}{(n-1)!^2}\frac{4^2}{(2n-1)(2n+1)^2}\frac{1}{2^{2n+1}}\int_0^1f^{(n)}(t)^2dt\\ &=\frac{1}{(n-1)!^2}\frac{8}{(4n^2-1)(2n+1)}\frac{1}{4^n}\int_0^1f^{(n)}(t)^2dt\\ &\leq\frac{1}{(n-1)!^2}\frac{8}{4n^2(2n+1)}\frac{1}{4^n}\int_0^1f^{(n)}(t)^2dt\\ &=\frac{2}{n!^2(2n+1)4^n}\int_0^1f^{(n)}(t)^2dt\\ \end{align*}

Unfortunately this is worse than what we wanted to show by a factor of $2$, perhaps one can improve one of the above estimates or I just made a mistake somewhere?