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I have difficulty understanding this proof of Shilov's book, Elementary Functional Analysis Elementary Functional Analysis by Georgi E. Shilov. P.35

Here is my trial to understand the proof. Given an $\epsilon > 0$ and a sequence $\{ y_n\}$ where $y_n(x) \rightarrow 0$ uniformally on $[a,c-\epsilon]$ and $y_n(x) \rightarrow 1$ uniformly on $[c+\epsilon,b]$ then, consider the distance between $y_m(x)$ and $y_n(x)$ and for sufficiently large $m,n$ we have $\rho^{p}(y_m(x),y_n(x)) = \int_{a}^{c-\epsilon} 0 dx +\int_{c-\epsilon}^{c+\epsilon} | y_m(x)-y_n(x)|^{p} dx + \int_{c+\epsilon}^{b}0 dx = \int_{c-\epsilon}^{c+\epsilon} | y_m(x)-y_n(x)|^{p} \leq \int_{c-\epsilon}^{c+\epsilon} 1 dx = 2\epsilon$ then indeed $\{y_n(x)\}$ is a cauchy sequence on $[a,b]$.

Question. Why $\rho^{p}(y_m(x),y_n(x)) \leq 4\epsilon$?

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    $\begingroup$ $y_m,y_n$ need not equal on $[a,c-\epsilon]$ (and $[c+\epsilon,b]$), so you need two $\epsilon$s to help you there. $\endgroup$ – user10354138 Nov 9 '18 at 9:29
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I got that in the proof he said that $y_n(x)$ is a sequence of functions taking values between $0$ and$1$,that is for each $n\in N$ and $x\in[a,b]$ we have $$0\leq y_n(x)\leq 1$$. therefore $$|y_n(x)-y_m(x)|\leq 1$$. Now since the sequence ${y_n(x)}$ is a converges sequence in $[a,c-e]$ and$[c+e,b]$ ,then there is a natural number $N$ such that for each $n,m\geq N$ and $x \in [a,c-e]$ or $x\in [c+e,b]$ we have $$|y_n(x)-y_m(x)| \lt ({\frac{e}{b-a}})^\frac 1p=T$$. Hence $$\int_{a}^{c-e}|y_n(x)-y_m(x)|^p \lt \int_{a}^{c-e}{T}^p=\int_{a}^{c-e}\frac{e}{b-a}=\frac{e}{b-a}((c-e)-a) \leq \frac{e}{b-a}(b-a)=e$$ Similarlly we have$$\int_{c+e}^{b}|y_n(x)-y_m(x)| \lt e$$. Hence $$\int_a^b |y_n(x)-y_m(x)| \lt e+2e+e=4e$$

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