The space $(C[a,b],rho_{p})$ is Incomplete proof











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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 $rho^{p}(y_m(x),y_n(x)) = int_{a}^{c-epsilon} 0 dx +int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} dx + int_{c+epsilon}^{b}0 dx = int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} leq int_{c-epsilon}^{c+epsilon} 1 dx = 2epsilon$ then indeed ${y_n(x)}$ is a cauchy sequence on $[a,b]$.



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






real-analysis functional-analysis analysis proof-explanation banach-spaces







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  • 2




    $y_m,y_n$ need not equal on $[a,c-epsilon]$ (and $[c+epsilon,b]$), so you need two $epsilon$s to help you there.
    – user10354138
    Nov 9 at 9:29















up vote
2
down vote

favorite












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 $rho^{p}(y_m(x),y_n(x)) = int_{a}^{c-epsilon} 0 dx +int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} dx + int_{c+epsilon}^{b}0 dx = int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} leq int_{c-epsilon}^{c+epsilon} 1 dx = 2epsilon$ then indeed ${y_n(x)}$ is a cauchy sequence on $[a,b]$.



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






real-analysis functional-analysis analysis proof-explanation banach-spaces







share|cite|improve this question




















  • 2




    $y_m,y_n$ need not equal on $[a,c-epsilon]$ (and $[c+epsilon,b]$), so you need two $epsilon$s to help you there.
    – user10354138
    Nov 9 at 9:29













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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 $rho^{p}(y_m(x),y_n(x)) = int_{a}^{c-epsilon} 0 dx +int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} dx + int_{c+epsilon}^{b}0 dx = int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} leq int_{c-epsilon}^{c+epsilon} 1 dx = 2epsilon$ then indeed ${y_n(x)}$ is a cauchy sequence on $[a,b]$.



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






real-analysis functional-analysis analysis proof-explanation banach-spaces







share|cite|improve this question















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 $rho^{p}(y_m(x),y_n(x)) = int_{a}^{c-epsilon} 0 dx +int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} dx + int_{c+epsilon}^{b}0 dx = int_{c-epsilon}^{c+epsilon} | y_m(x)-y_n(x)|^{p} leq int_{c-epsilon}^{c+epsilon} 1 dx = 2epsilon$ then indeed ${y_n(x)}$ is a cauchy sequence on $[a,b]$.



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






real-analysis functional-analysis analysis proof-explanation banach-spaces




real-analysis functional-analysis analysis proof-explanation banach-spaces






share|cite|improve this question















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edited Nov 21 at 18:48

























asked Nov 9 at 9:24









HybridAlien

2008




2008








  • 2




    $y_m,y_n$ need not equal on $[a,c-epsilon]$ (and $[c+epsilon,b]$), so you need two $epsilon$s to help you there.
    – user10354138
    Nov 9 at 9:29














  • 2




    $y_m,y_n$ need not equal on $[a,c-epsilon]$ (and $[c+epsilon,b]$), so you need two $epsilon$s to help you there.
    – user10354138
    Nov 9 at 9:29








2




2




$y_m,y_n$ need not equal on $[a,c-epsilon]$ (and $[c+epsilon,b]$), so you need two $epsilon$s to help you there.
– user10354138
Nov 9 at 9:29




$y_m,y_n$ need not equal on $[a,c-epsilon]$ (and $[c+epsilon,b]$), so you need two $epsilon$s to help you there.
– user10354138
Nov 9 at 9:29










1 Answer
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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 $nin N$ and $xin[a,b]$ we have $$0leq y_n(x)leq 1$$.
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 $n,mgeq N$ and $x in [a,c-e]$ or $xin [c+e,b]$ 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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    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 $nin N$ and $xin[a,b]$ we have $$0leq y_n(x)leq 1$$.
    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 $n,mgeq N$ and $x in [a,c-e]$ or $xin [c+e,b]$ 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
    $$int_a^b |y_n(x)-y_m(x)| lt e+2e+e=4e$$






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      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 $nin N$ and $xin[a,b]$ we have $$0leq y_n(x)leq 1$$.
      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 $n,mgeq N$ and $x in [a,c-e]$ or $xin [c+e,b]$ 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
      $$int_a^b |y_n(x)-y_m(x)| lt e+2e+e=4e$$






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        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 $nin N$ and $xin[a,b]$ we have $$0leq y_n(x)leq 1$$.
        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 $n,mgeq N$ and $x in [a,c-e]$ or $xin [c+e,b]$ 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
        $$int_a^b |y_n(x)-y_m(x)| lt e+2e+e=4e$$






        share|cite|improve this answer












        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 $nin N$ and $xin[a,b]$ we have $$0leq y_n(x)leq 1$$.
        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 $n,mgeq N$ and $x in [a,c-e]$ or $xin [c+e,b]$ 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
        $$int_a^b |y_n(x)-y_m(x)| lt e+2e+e=4e$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 9 at 10:49









        ehsan eskandari

        985




        985






























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