Prove that $phi_{h}(x) = frac{1}{2h}int_{x-h}^{x+h}f(t)dt in L(mathbb{R})$
Problem. Given $f in L(mathbb{R})$, let
$$phi_{h}(x) = frac{1}{2h}int_{x-h}^{x+h}f(t)dt,quad h>0.$$
Prove that $phi in L(mathbb{R})$ e $displaystyle int_{mathbb{R}}|phi_{h}(x)|dx leq Vert f Vert_{1}.$
My attempt.
begin{eqnarray*}
int_{mathbb{R}}|phi_{h}(x)| & leq & frac{1}{2h}int_{mathbb{R}}int_{[x-h,x+h]}|f(t)|dtdx\
& = & frac{1}{2h}underbrace{int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dtdx}_{text{Fubini-Tonelli}}\
& = & frac{1}{2h}int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dxdt\
& = & frac{1}{2h}int_{mathbb{R}}|f(t)|underbrace{m([x-h,x+h])}_{mtext{ is the Lebesgue measure}}dt\
& = & int_{mathbb{R}}|f(t)|dt\
& = & Vert f Vert_{1}
end{eqnarray*}
Is this correct?
measure-theory lebesgue-measure
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
add a comment |
Problem. Given $f in L(mathbb{R})$, let
$$phi_{h}(x) = frac{1}{2h}int_{x-h}^{x+h}f(t)dt,quad h>0.$$
Prove that $phi in L(mathbb{R})$ e $displaystyle int_{mathbb{R}}|phi_{h}(x)|dx leq Vert f Vert_{1}.$
My attempt.
begin{eqnarray*}
int_{mathbb{R}}|phi_{h}(x)| & leq & frac{1}{2h}int_{mathbb{R}}int_{[x-h,x+h]}|f(t)|dtdx\
& = & frac{1}{2h}underbrace{int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dtdx}_{text{Fubini-Tonelli}}\
& = & frac{1}{2h}int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dxdt\
& = & frac{1}{2h}int_{mathbb{R}}|f(t)|underbrace{m([x-h,x+h])}_{mtext{ is the Lebesgue measure}}dt\
& = & int_{mathbb{R}}|f(t)|dt\
& = & Vert f Vert_{1}
end{eqnarray*}
Is this correct?
measure-theory lebesgue-measure
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
2
no. first equality should be an inequality
– mathworker21
Dec 10 '18 at 4:58
2
But other than that, it is correct.
– copper.hat
Dec 10 '18 at 5:02
Yes! Of course.. its a typo. Thank you
– Lucas Corrêa
Dec 10 '18 at 5:02
add a comment |
Problem. Given $f in L(mathbb{R})$, let
$$phi_{h}(x) = frac{1}{2h}int_{x-h}^{x+h}f(t)dt,quad h>0.$$
Prove that $phi in L(mathbb{R})$ e $displaystyle int_{mathbb{R}}|phi_{h}(x)|dx leq Vert f Vert_{1}.$
My attempt.
begin{eqnarray*}
int_{mathbb{R}}|phi_{h}(x)| & leq & frac{1}{2h}int_{mathbb{R}}int_{[x-h,x+h]}|f(t)|dtdx\
& = & frac{1}{2h}underbrace{int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dtdx}_{text{Fubini-Tonelli}}\
& = & frac{1}{2h}int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dxdt\
& = & frac{1}{2h}int_{mathbb{R}}|f(t)|underbrace{m([x-h,x+h])}_{mtext{ is the Lebesgue measure}}dt\
& = & int_{mathbb{R}}|f(t)|dt\
& = & Vert f Vert_{1}
end{eqnarray*}
Is this correct?
measure-theory lebesgue-measure
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
Problem. Given $f in L(mathbb{R})$, let
$$phi_{h}(x) = frac{1}{2h}int_{x-h}^{x+h}f(t)dt,quad h>0.$$
Prove that $phi in L(mathbb{R})$ e $displaystyle int_{mathbb{R}}|phi_{h}(x)|dx leq Vert f Vert_{1}.$
My attempt.
begin{eqnarray*}
int_{mathbb{R}}|phi_{h}(x)| & leq & frac{1}{2h}int_{mathbb{R}}int_{[x-h,x+h]}|f(t)|dtdx\
& = & frac{1}{2h}underbrace{int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dtdx}_{text{Fubini-Tonelli}}\
& = & frac{1}{2h}int_{mathbb{R}}int_{mathbb{R}}|f(t)|chi_{[x-h,x+h]}dxdt\
& = & frac{1}{2h}int_{mathbb{R}}|f(t)|underbrace{m([x-h,x+h])}_{mtext{ is the Lebesgue measure}}dt\
& = & int_{mathbb{R}}|f(t)|dt\
& = & Vert f Vert_{1}
end{eqnarray*}
Is this correct?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
edited Dec 10 '18 at 5:02
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
asked Dec 10 '18 at 4:57
Lucas Corrêa
1,4471321
1,4471321
2
no. first equality should be an inequality
– mathworker21
Dec 10 '18 at 4:58
2
But other than that, it is correct.
– copper.hat
Dec 10 '18 at 5:02
Yes! Of course.. its a typo. Thank you
– Lucas Corrêa
Dec 10 '18 at 5:02
add a comment |
2
no. first equality should be an inequality
– mathworker21
Dec 10 '18 at 4:58
2
But other than that, it is correct.
– copper.hat
Dec 10 '18 at 5:02
Yes! Of course.. its a typo. Thank you
– Lucas Corrêa
Dec 10 '18 at 5:02
2
2
no. first equality should be an inequality
– mathworker21
Dec 10 '18 at 4:58
no. first equality should be an inequality
– mathworker21
Dec 10 '18 at 4:58
2
2
But other than that, it is correct.
– copper.hat
Dec 10 '18 at 5:02
But other than that, it is correct.
– copper.hat
Dec 10 '18 at 5:02
Yes! Of course.. its a typo. Thank you
– Lucas Corrêa
Dec 10 '18 at 5:02
Yes! Of course.. its a typo. Thank you
– Lucas Corrêa
Dec 10 '18 at 5:02
add a comment |
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2
no. first equality should be an inequality
– mathworker21
Dec 10 '18 at 4:58
2
But other than that, it is correct.
– copper.hat
Dec 10 '18 at 5:02
Yes! Of course.. its a typo. Thank you
– Lucas Corrêa
Dec 10 '18 at 5:02