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.



Steklov means (A paper by H.Johnen and K. Scherer)
Ok, I think I figured it out.


We can notice that the $|(-1)^r|=1$,so the sum 


$$\bigg| \sum\_{k=0}^r (-1)^{r-k}\ldots \bigg| = \bigg| \sum\_{k=0}^r (-1)^{-k}\ldots \bigg| = \bigg| (-1)^r f(x) + \sum\_{k=0}^r (-1)^{r-k} \ldots \bigg|$$


That solves my problem.


Embarrassing... :-D


