Lebesgue space inclusion on half real line
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Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.
Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?
real-analysis integration lebesgue-integral lebesgue-measure
asked Nov 22 at 16:36
sleepingrabbit
1317
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up vote
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Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.
Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?
real-analysis integration lebesgue-integral lebesgue-measure
asked Nov 22 at 16:36
sleepingrabbit
1317
I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.
Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?
real-analysis integration lebesgue-integral lebesgue-measure
asked Nov 22 at 16:36
sleepingrabbit
1317
Assume that $f in L^1 ([0,infty))$ is a real continuous function on $[0,infty)$.
Can we then conclude that $f in L^p ([0,infty))$ for $p in [1, infty)$?
real-analysis integration lebesgue-integral lebesgue-measure
real-analysis integration lebesgue-integral lebesgue-measure
asked Nov 22 at 16:36
sleepingrabbit
1317
asked Nov 22 at 16:36
sleepingrabbit
1317
asked Nov 22 at 16:36
sleepingrabbit
1317
asked Nov 22 at 16:36
sleepingrabbit
1317
asked Nov 22 at 16:36
sleepingrabbit
1317
1317
I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45
add a comment |
I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45
I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45
I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45
add a comment |
1 Answer
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This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
$$
int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
$$
Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
$$
int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
$$
This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.
answered Nov 22 at 16:55
Daniel
1,516210
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
$$
int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
$$
Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
$$
int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
$$
This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.
answered Nov 22 at 16:55
Daniel
1,516210
add a comment |
up vote
2
down vote
accepted
This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
$$
int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
$$
Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
$$
int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
$$
This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.
answered Nov 22 at 16:55
Daniel
1,516210
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
$$
int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
$$
Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
$$
int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
$$
This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.
answered Nov 22 at 16:55
Daniel
1,516210
This is false. To build a counterexample, consider functions $g_n(x)$ that is non-zero only on $(n, n + 2n^{-3})$, piecewise linear, with $g_n(n) = g_n(n+2n^{-3}) = 0$ and $g_n(n+n^{-3}) = n$. Then,
$$
int g_n(x) dx = frac12 n cdot 2n^{-3} = n^{-2}.
$$
Since for $n geq 2$ functions $g_n$ have disjoint support, if we define $f = sum_{n geq 2} g_n$ then $f in L^1([0, infty))$. However, if we take $p = 2$, we have that
$$
int g_n(x)^2 dx geq int_{n + frac12 n^{-3}}^{n + frac32 n^{-3}} (frac12 n)^2 dx = frac14 n^{-1}.
$$
This implies $int f^2 geq frac14 sum_{n geq 2} n^{-1} = infty$.
answered Nov 22 at 16:55
Daniel
1,516210
edited Nov 22 at 17:29
answered Nov 22 at 16:55
Daniel
1,516210
answered Nov 22 at 16:55
Daniel
1,516210
answered Nov 22 at 16:55
Daniel
1,516210
1,516210
add a comment |
add a comment |
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I guess $f(x) = 1$ is a real continuous function on $[0,infty)$. Is it in $L^p$?
– GEdgar
Nov 22 at 16:45