Elementary method to prove that $|\overline B_{2n}(x)|\le|B_{2n}|$?

Question

I am currently trying to estimate the error term in Euler-Maclaurin summation formula and need to establish an upper bound for the periodic Bernoulli polynomial $\overline B_{2n}(x)=B_{2n}(x-\lfloor x\rfloor)$. I know that it is obvious that the following Fourier expansion implies $|\overline B_{2n}(x)|\le|B_{2n}|$.

$$ B_{2n}(x)={2(-1)^{n+1}\over(2\pi)^{2n}}\sum_{k=1}^\infty{\cos(2\pi kx)\over k^{2n}} $$

But I wonder if it is possible to establish this bound by merely elementary methods (e.g. induction, real-variable calculus, etc.).

Answer 1

A weaker bound.

Notation: For clarity, Bernoulli polynomials will be written with a regular italic font $B$ whereas the Bernoulli numbers will be written with a bolded $\boldsymbol{B}$.

The $n=0$ and $n=1$ cases can both be checked individually very easily, so we'll now assume $n\geq 2$. The Bernoulli polynomials have a well known differential recurrence relation $${B_n}'=nB_{n-1}$$ Amazingly, this formula extends to the periodic Bernoulli polynomials as well: $${\overline{B}_n}'=n\overline B_{n-1}$$ We can shuffle the index, integrate both sides of the equation, and use the special value $\overline{B}_n(0)=B_n(0)=\boldsymbol B_n$ to get

$$\overline B_{n+1}(x)=(n+1)\int_0^x \overline B_n(s)\mathrm ds+B_{n+1}$$ Applying this integral recursion twice and using the fact that when $n\geq 2$, $\boldsymbol B_{2n+1}=0$, we get $$\overline B_{2n+2}(x)=\int\limits_0^x\int\limits_0^s\overline B_{2n}(t)\mathrm dt\mathrm ds+\boldsymbol B_{2n}\cdot x$$ It is important to note that since the periodic Bernoulli polynomials are, well, periodic, we can assume WLOG that $x\in[0,1]$. Let's now take the absolute value on both sides, and using the triangle inequality, $$\left|\overline B_{2n+2}(x)\right|=\left|\int\limits_0^x\int\limits_0^s\overline B_{2n}(t)\mathrm dt\mathrm ds+\boldsymbol B_{2n}\cdot x\right|\leq \left|\int\limits_0^x\int\limits_0^s\overline B_{2n}(t)\mathrm dt\mathrm ds\right|+\left|\boldsymbol B_{2n}\cdot x\right|$$ And now, using the integral triangle inequality (proof here ) and the fact that $|x|\leq 1$ we get $$\left|\overline B_{2n+2}(x)\right|\leq\int\limits_0^x\int\limits_0^s\left|\overline B_{2n}(t)\right|\mathrm dt\mathrm ds+\left|\boldsymbol B_{2n}\right|$$

Induction!

Assume the hypothesis is true for $\overline B_{2n}$. Then, $$\int_0^s |\overline B_{2n}(t)|\mathrm{d}t\leq\int_0^s |\boldsymbol B_{2n}|\mathrm dt=|\boldsymbol{B}_{2n}|s$$ Thus, $$\int\limits_0^x\int\limits_0^s\left|\overline B_{2n}(t)\right|\mathrm dt\mathrm ds\leq\int_0^x |\boldsymbol B_{2n}|s~\mathrm ds=|\boldsymbol B_{2n}|\frac{x^2}{2}$$ Since $x\in[0,1]$, we have $$\int\limits_0^x\int\limits_0^s\left|\overline B_{2n}(t)\right|\mathrm dt\mathrm ds\leq\int_0^x |\boldsymbol B_{2n}|s~\mathrm ds\leq \frac{|\boldsymbol B_{2n}|}{2}$$ So in all we have $$\left|\overline B_{2n+2}(x)\right|\leq \frac{3}{2}\left|\boldsymbol B_{2n}\right|$$

Can anyone improve this approach?

Answer 2

Still, another weaker result

This is a translation from an answer to the same question on Zhihu.

Well, the generating function for Bernoulli polynomials is sufficient for us to establish that bound:

$$ {xe^{tx}\over e^x-1}=\sum_{n=0}^\infty{x^n\over n!}B_n(t) $$

If we substitute that $t$ with $1-t$, we have

$$ {xe^{(1-t)x}\over e^x-1}={(-x)e^{t(-x)}\over e^{-x}-1} $$

Comparing the coefficients, we can see that

$$ B_n(1-t)=(-1)^nB_n(t) $$

In particular, the above equation implies $B_{2n-1}(1-t)=-B_{2n-1}(t)$ for all $n\ge2$. As a result, we see that for all $n\ge1$, $\{0,1/2,1\}\subset\{t\in[0,1]:B_{2n-1}(t)=0\}$.

To show that the $B_{2n-1}(t)$ does not have other zeros, let's first assume that there exists $c\in(0,1/2)$ such that $B_{2n-1}(c)=0$. This implies that $B_{2n-1}(t)$ has at least 3 stationary points in $[0,1]$, so $B_{2n-2}(t)$ has at least 3 zeros. Using similar reasoning, we observe that $B_{2n-3}(t)$ has at least 4 zeros in $[0,1]$. Propagating like this, we observe that the case for $B_{2}(t)$ will result in contradiction.

Based on the above conclusion, we see that $B_{2n-1}(t)$ does not change sign in $[0,1/2]$ and in $[1/2,1]$ respectively. This indicates the following relation:

$$ |B_{2n}(x)|\le\max\{|B_{2n}|,|B_{2n}(1/2)|\} $$