Prove integral inequality including n-order derivative 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?
 A: 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
$$
\int_0^1 \min(t, 1-t)^{2n} \, dt = 2 \int_0^{1/2} t^{2n} \, dt
= \frac{1}{(2n+1)4^n} \, .
$$
A: 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?
