It's not clear to me how this passage has been performed (Yosida, Functional Analysis page 9).
$$ \bigg|\sum_{|p-nx| > \delta n} (f(x) - f(p/n))r_p(x)\bigg| \leq 2 M \sum_{|p-nx| > \delta n} r_p(x) \leq \dfrac{2M}{n^2\delta^2} \sum_{p = 0}^n (p - nx)^2 r_p(x) $$
Where what we are doing is to prove the Weiestrass' Polynomial Approximation Theory: $f(x)$ is a real valued function, continuous on $[0, 1]$ and there exists a polynomial $P_n(x)$ which converges to $f(x)$ as $n \to +\infty$ uniformly on $[0, 1]$.
We took
$$P_n(x) = \sum_{p = 0}^n \binom{n}{k} f(p/n) x^p (1-x)^{n-p}$$
We got those identities from binomial expansion, calling $r_p(x) = \binom{n}{k} x^p (1-x)^{n-p}$:
$$\sum_{p =0}^n r_p(x) = 1$$ $$\sum_{p =0}^n p r_p(x) = nx$$ $$\sum_{p =0}^n p(p-1) r_p(x) = n(n-1)x^2$$
(Those are all clear, I proved them).
Also:
$$\sum_{p =0}^n (p-nx)^2 r_p(x) = nx(1-x)$$
Now we assumed $|f(x)| \leq M < +\infty$ on $[0, 1]$ and by uniform continuity $|f(x) - f(x')| < \epsilon$ when $|x - x'| \leq \delta$.
Here I understood $x' = p/n$.
Then by using the definition he wrote
$$\bigg| f(x) - P_n(x)\bigg| = \bigg|\sum_{p = 0}^n (f(x) - f(p/n))r_p(x)\bigg| \leq \bigg| \sum_{|p-nx|\leq \delta n}\bigg| + \bigg| \sum_{|p-nx| > \delta n}\bigg|$$
The first term I understood the procedure (the term with the sum $\leq \delta n$).
The other term I did not understand the passage, so the passage I'm asking for is this:
$$ \bigg|\sum_{|p-nx| > \delta n} (f(x) - f(p/n))r_p(x)\bigg| \leq 2 M \sum_{|p-nx| > \delta n} r_p(x) \leq \dfrac{2M}{n^2\delta^2} \sum_{p = 0}^n (p - nx)^2 r_p(x) $$
Thoughts
- First of all I did not understand where $2M$ comes from. I mean ok $|f(x)| \leq M$. But we have $|f(x) - f(x')|$ so is this just a rough estimation? Something like
$$|f(x) - f(x')| \leq |f(x)| - |f(x')| \leq M + M = 2M$$
- The most cryptic part is how it goes from $\sum_{|p-nx| > \delta n}$ to $\sum_{p = 0}^n$
I thought of a shift like
$$|p - nx| > \delta n \rightarrow (p - nx)^2 > \delta^2 n^2 \rightarrow (p - nx)^2 - \delta^2 n^2 > 0$$
Hence from $0$ to $n$, by shifting the summation index. This will mean something like $p' = (p - nx)^2 - \delta^2 n^2$, whence it would change the $r_p(x)$ factor $\mapsto r_{p'}(x)$
But still I don't get how $(p-nx)^2$ pops out in the summation term.
After that it's all clear since then we would have
$$(\ldots) \leq \dfrac{2M}{n^2\delta^2} \sum_{p = 0}^n (p - nx)^2 r_p(x) = \dfrac{2Mx(1-x)}{n\delta^2} \to 0$$
as $n\to +\infty$. Thank you!
Instead of seeing it as $(p - nx)^2 - \delta^2 n^2 > 0$, you should see it as $\displaystyle\frac{(p - nx)^2}{\delta^2 n^2} > 1$, that is, you're seeing $r_p(x)$ as $1 \cdot r_p(x)$ and artificially introducing each $\displaystyle\frac{(p - nx)^2}{\delta^2 n^2}$ thanks to the inequality, and the denominator doesn't depend on $p$ so you extract it.
The subtle part is that you do that for the $p$s such that the fraction is over $1$, but then also artificially add $\displaystyle\frac{(p - nx)^2}{\delta^2 n^2} r_p(x)$ for $p$s where the fraction is less than $1$ by saying that each term is greater than $0$.
To sum it up, you do this: $$\begin{split}\sum_{|p - nx| > \delta n} r_p(x) &= \sum_{|p - nx| > \delta n} 1 \cdot r_p(x) + \sum_{|p - nx| \leq \delta n} 0 \\ &\leq \sum_{|p - nx| > \delta n} \frac{(p - nx)^2}{\delta^2 n^2} r_p(x) + \sum_{|p - nx| \leq \delta n} \frac{(p - nx)^2}{\delta^2 n^2} r_p(x)\\ &\leq \sum_{p = 0}^n \frac{(p - nx)^2}{\delta^2 n^2} r_p(x)\end{split}$$
The squaring is for the purpose of the proof, and the $2M$ is also indeed a rough estimation in the way you described.