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There are no bounded linear functional on $L^p(mathbb{R})$ if $0<p<1.$
Consider $L^p(mathbb{R})$ where $0<p<1$.
Why there are no bounded linear functional on $L^p(mathbb{R})$?
i.e.
If $l$ is linear functional $l:L^p(mathbb{R})to mathbb{C}$ such that $|l(f)|leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$.
Why?
Hint: Let $F(x)=l(mathcal{X}_{[0,x]})$, and consider $F(x)-F(y)$
functional-analysis measure-theory lp-spaces
asked Nov 21 '18 at 5:24
eraldcoil
363111
add a comment |
Consider $L^p(mathbb{R})$ where $0<p<1$.
Why there are no bounded linear functional on $L^p(mathbb{R})$?
i.e.
If $l$ is linear functional $l:L^p(mathbb{R})to mathbb{C}$ such that $|l(f)|leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$.
Why?
Hint: Let $F(x)=l(mathcal{X}_{[0,x]})$, and consider $F(x)-F(y)$
functional-analysis measure-theory lp-spaces
asked Nov 21 '18 at 5:24
eraldcoil
363111
add a comment |
Consider $L^p(mathbb{R})$ where $0<p<1$.
Why there are no bounded linear functional on $L^p(mathbb{R})$?
i.e.
If $l$ is linear functional $l:L^p(mathbb{R})to mathbb{C}$ such that $|l(f)|leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$.
Why?
Hint: Let $F(x)=l(mathcal{X}_{[0,x]})$, and consider $F(x)-F(y)$
functional-analysis measure-theory lp-spaces
asked Nov 21 '18 at 5:24
eraldcoil
363111
Consider $L^p(mathbb{R})$ where $0<p<1$.
Why there are no bounded linear functional on $L^p(mathbb{R})$?
i.e.
If $l$ is linear functional $l:L^p(mathbb{R})to mathbb{C}$ such that $|l(f)|leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$.
Why?
Hint: Let $F(x)=l(mathcal{X}_{[0,x]})$, and consider $F(x)-F(y)$
functional-analysis measure-theory lp-spaces
functional-analysis measure-theory lp-spaces
asked Nov 21 '18 at 5:24
eraldcoil
363111
asked Nov 21 '18 at 5:24
eraldcoil
363111
edited Nov 21 '18 at 6:25
asked Nov 21 '18 at 5:24
eraldcoil
363111
asked Nov 21 '18 at 5:24
eraldcoil
363111
asked Nov 21 '18 at 5:24
eraldcoil
363111
363111
add a comment |
add a comment |
1 Answer
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This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(mathbb{R})$ is only a metric space with invariant metric given by
$$|f|_p := int |f|^p d lambda.$$
If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f in B_delta(0)$ and some $delta >0$. By rescaling and using linearity, we see that $$|l(f)| le M |f|_p^{1/p}$$
with $M= delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,ldots,I_n$ with lenght $(b-a)/n$. Then we have
$$|l(1_{[a,b]})| le sum_{k=1}^n |l(1_{I_i})| le nM |l(1_{I_i})|_p = (b-a)^{1/p} n^{1-1/p}.$$
Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n rightarrow infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.
answered Nov 22 '18 at 15:47
p4sch
4,770217
add a comment |
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1 Answer
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1 Answer
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active
oldest
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This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(mathbb{R})$ is only a metric space with invariant metric given by
$$|f|_p := int |f|^p d lambda.$$
If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f in B_delta(0)$ and some $delta >0$. By rescaling and using linearity, we see that $$|l(f)| le M |f|_p^{1/p}$$
with $M= delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,ldots,I_n$ with lenght $(b-a)/n$. Then we have
$$|l(1_{[a,b]})| le sum_{k=1}^n |l(1_{I_i})| le nM |l(1_{I_i})|_p = (b-a)^{1/p} n^{1-1/p}.$$
Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n rightarrow infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.
answered Nov 22 '18 at 15:47
p4sch
4,770217
add a comment |
This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(mathbb{R})$ is only a metric space with invariant metric given by
$$|f|_p := int |f|^p d lambda.$$
If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f in B_delta(0)$ and some $delta >0$. By rescaling and using linearity, we see that $$|l(f)| le M |f|_p^{1/p}$$
with $M= delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,ldots,I_n$ with lenght $(b-a)/n$. Then we have
$$|l(1_{[a,b]})| le sum_{k=1}^n |l(1_{I_i})| le nM |l(1_{I_i})|_p = (b-a)^{1/p} n^{1-1/p}.$$
Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n rightarrow infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.
answered Nov 22 '18 at 15:47
p4sch
4,770217
add a comment |
This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(mathbb{R})$ is only a metric space with invariant metric given by
$$|f|_p := int |f|^p d lambda.$$
If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f in B_delta(0)$ and some $delta >0$. By rescaling and using linearity, we see that $$|l(f)| le M |f|_p^{1/p}$$
with $M= delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,ldots,I_n$ with lenght $(b-a)/n$. Then we have
$$|l(1_{[a,b]})| le sum_{k=1}^n |l(1_{I_i})| le nM |l(1_{I_i})|_p = (b-a)^{1/p} n^{1-1/p}.$$
Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n rightarrow infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.
answered Nov 22 '18 at 15:47
p4sch
4,770217
This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(mathbb{R})$ is only a metric space with invariant metric given by
$$|f|_p := int |f|^p d lambda.$$
If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f in B_delta(0)$ and some $delta >0$. By rescaling and using linearity, we see that $$|l(f)| le M |f|_p^{1/p}$$
with $M= delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,ldots,I_n$ with lenght $(b-a)/n$. Then we have
$$|l(1_{[a,b]})| le sum_{k=1}^n |l(1_{I_i})| le nM |l(1_{I_i})|_p = (b-a)^{1/p} n^{1-1/p}.$$
Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n rightarrow infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.
answered Nov 22 '18 at 15:47
p4sch
4,770217
answered Nov 22 '18 at 15:47
p4sch
4,770217
answered Nov 22 '18 at 15:47
p4sch
4,770217
answered Nov 22 '18 at 15:47
p4sch
4,770217
4,770217
add a comment |
add a comment |
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