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

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