I asked this question in MSE about a month ago, and didn't get a reply as of yet.
I'll copy and paste this question here, hopefully to get some response.
I am reading the following paper: http://www.kryakin.org/k/butzer_johnen_k-func.pdf called: ON THE EQUIVALENCE OF THE K-FUNCTIONAL AND MODULI OF CONTINUITY AND SOME APPLICATIONS by H.Johnen and K. Scherer
on page 124, in the proof of lemma 3, they write the Steklov-means of $f(x)$, as in $$g_s(x) = -\sum_{k=1}^r (-1)^{r-k}{r\choose k} \int_0^1 \ldots \int_0^1 f(x+k\cdot \sum_{i=1}^n \sum_{j=1}^r s\sigma_{ij}e_i)\prod_{i=1}^n \prod_{j=1}^r d\sigma_{ij}$$
And we recall that: $\Delta^r(h)f(x) = \sum_{k=0}^r (-1)^{r-k} {r\choose k} f(x+kh)$.
I don't see how did they deduce the inequality below equation (2.9), i.e: $$\| f-g_s \|_{L^p(V)} \le \int_0^1 \ldots \int_0^1 \| \Delta^r(s\sum_{i=1}^n \sum_{j=1}^r \sigma_{ij}e_i)f \|_{L^p(V)}\prod_{i=1}^n \prod_{j=1}^r d\sigma_{ij}$$
So I want to show this inequality but I got stuck, if I am not wrong we should have a $k=0$ term in the $r$-difference operator;
So if I write it as: $$\bigg| (f-g_s)(x) \bigg| = \bigg| f(x)+ \sum_{k=1}^r (-1)^{r-k} {r\choose k} \int_0^1 \ldots \int_0^1 f(x+k\cdot \sum_{i=1}^n \sum_{j=1}^r s\sigma_{ij}e_i)\prod_{i=1}^n \prod_{j=1}^r d\sigma_{ij} \bigg|$$
Now, if I am not wrong for us to get in the RHS the following term: $\bigg| \sum_{k=0}^r (-1)^{r-k} {r\choose k}\int_0^1 \ldots \int_0^1 f(x+k\cdot \sum_{i=1}^n \sum_{j=1}^r s\sigma_{ij}e_i)\prod_{i=1}^n\prod_{j=1}^r d\sigma_{ij} \bigg|$, we should have $(-1)^rf(x)$ instead of $f(x)$, how do they bypass this?
I think I am missing something in this easy derivation, but I don't see it.
Your help is appreciated.
