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?


2 Answers 2


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 $$ \int_0^1 \min(t, 1-t)^{2n} \, dt = 2 \int_0^{1/2} t^{2n} \, dt = \frac{1}{(2n+1)4^n} \, . $$


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?

  • $\begingroup$ I hope there is no too much overlap between my solution and yours, I had developed that independently in the meantime. $\endgroup$
    – Martin R
    Aug 17, 2021 at 14:31
  • $\begingroup$ @MartinR All good, my solution wasn't complete anyways. It seems like the only non-trivial overlap is the use of the fact that the integral of $f$ is the integral of the remainder term, but this is what the OP had already noticed anyways. It was really your trick interchanging the order of integration by splitting up the integral that made it work out. $\endgroup$
    – leoli1
    Aug 17, 2021 at 20:16

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