# An other total-variation inequality of mine

Given differentiable, continuous $$f\left ( x \right )$$ on the interval $$\left [ 0, 1 \right ]$$ so that $$\int_{0}^{1}f\left ( x \right ){\rm d}x= 0.$$ Prove that $$\left | \int_{0}^{1}xf\left ( x \right ){\rm d}x \right |\leq \frac{1}{12}\max\left | {f}'\left ( x \right ) \right |$$

I think I should transform the constant $$1/12$$ into an integral like $$k\int_{0}^{1}x^{2}{\rm d}x,$$ but $$k$$ is very unusual, I need to your helps, even an example of $$f\left ( x \right )$$ so that $$\int_{0}^{1}f\left ( x \right ){\rm d}x= 0$$ in order to know what I must do with the constant. Thanks a real lot.

$$\int_0^1 xf(x)\,dx=\int_0^1\left(x-\frac12\right)f(x)\,dx=\underbrace{-\frac12 x(1-x)f(x)\Bigg|_0^1}_{=0}+\frac12\int_0^1 x(1-x)f'(x)\,dx.$$ Hence, if $$\displaystyle M=\sup_{x\in(0,1)}|f'(x)|$$, then $$\displaystyle\left|\int_0^1 xf(x)\,dx\right|\leqslant\frac{M}{2}\int_0^1 x(1-x)\,dx=\frac{M}{12}$$.
Since $$\int_0^1f(x)\mathrm{d}x=0$$, by integration by parts, \begin{align*} \int_0^1xf(x)\mathrm{d}x&=\int_0^1(x+C)f(x)\mathrm{d}x\\ &=\Big(\frac{1}{2}x^2+Cx\Big)f(x)\bigg|_0^1-\int_0^1\Big(\frac{1}{2}x^2+Cx\Big)f'(x)\mathrm{d}x\\ &=\Big(\frac{1}{2}+C\Big)f(1)-\int_0^1\Big(\frac{1}{2}x^2+Cx\Big)f'(x)\mathrm{d}x \end{align*} Since $$f(1)$$ is unknown, so we let $$C=-\frac{1}{2}$$, we get $$\int_0^1xf(x)\mathrm{d}x=-\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)f'(x)\mathrm{d}x$$ Therefore, \begin{align*} \left|\int_0^1xf(x)\mathrm{d}x\right| &=\left|-\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)f'(x)\mathrm{d}x\right|\\ &=\left|-f'(\xi)\int_0^1\Big(\frac{1}{2}x^2-\frac{1}{2}x\Big)\mathrm{d}x\right|, \quad x\in(0,1)\\ &\leqslant\frac{1}{12}\max_{x\in[0,1]}|f'(x)| \end{align*}