supremum of $\int |\int f(x)-f(y)\,dy| \,dx $. Let $A_k$, $k\in \mathbb{N}$, be the family of $C^\infty([0,1])$ functions defined by
$$
A_k=\{||f^{(j)}||_{\infty}\le 1,\;\;0\le j \le k \}
$$
where $||\cdot||_{\infty}$ denotes the supremum norm over $[0,1]$ and $f^{(k)}$ is the k-th derivative of $f$. 
We define
$$
\sigma_k=\sup \left\{\int_0^1\left|\int_0^1f(x)-f(y)\,dy\right|\,dx ,\;\; f\in A_k\right\}
$$
What is the value of $\sigma_k$? 
Edit:
I think $\sigma_0=1$. I believe $\sigma_k= \frac{1}{4}$ for all $k>0$ (with $f=x$ the supremum is reached).
It is not homework and is not related to my work. Just curious. I hope you find interesting too. (((The problem turned out to be uninteresting.))) Thanks.
 A: Your conjecture is correct.
Let us note that we can rewrite your expression as 
$$\int_0^1 \left|\int_0^1 f(x)-f(y)\,dy\right|\,dx = \int_0^1 |f(x) - \bar{f}|\,dx$$ where $\bar{f} = \int_0^1 f(y)\,dy$.
For $k=0$, the Cauchy-Schwarz inequality gives
$$\begin{align*}
\left(\int_0^1 |f(x)-\bar{f}|\,dx\right)^2 &\le \int_0^1 (f(x)-\bar{f})^2\,dx \\
&= \int_0^1 f(x)^2\,dx - 2 \bar{f} \int_0^1 f(x)\,dx + \bar{f}^2 \\
&= \int_0^1 f(x)^2\,dx - \bar{f}^2 \\
&\le \int_0^1 f(x)^2\,dx \le 1.\end{align*}$$
Thus $\sigma_0 \le 1$.  We can show this is sharp by letting $f$ be a smooth function with $f=-1$ on $[0, \frac{1}{2}- \epsilon]$, and $f=1$ on $[\frac{1}{2} + \epsilon, 1]$.
The $k \ge 1$ case follows from results of the following paper:

Seng-Kee Chua and Richard L. Wheeden.  A note on sharp 1-dimensional Poincaré inequalities.  Proc. Amer. Math. Soc. 134(8) 2309-2316, 2006.  Open access PDF.

The authors prove, in our notation, that for any $p > 1$ and any Lipschitz $f$ defined on $[0,1]$, we have
$$\| f - \bar{f}\|_{L^1} \le \frac{1}{2} (1+p')^{-1/p'} \|f'\|_{L^p}$$
where presumably $1/p + 1/p' = 1$ (this would be the usual convention, though the authors do not seem to write it out).  For $f \in A_k$ we have $\|f'\|_{L^p} \le \|f'\|_\infty \le 1$, so letting $p \to \infty$ and $p' \to 1$ we get the bound $\|f - \bar{f}\|_{L^1} \le \frac{1}{4}$ as desired.  Your example $f(x) = x$ shows this is sharp.
