Mixing
$L^2$ is the only Hilbert space , Parallelogram Law and particular $f(t),g(t)$
$begingroup$
Let the space $C([0,1])$ and consider the norm $forall p in mathbb{N}$
$$forall f in C([0,1]),
||f ||_{L^P}= left ( int^1_0 |f(t)|^P dt right )^{frac{1}{p}} $$
knowing that when $p=2$, this norm is the norm induced by the innder product
$$forall f,g in C([0,1]), <f,g>=int^1_0 f(t)overline{g(t)} dx $$
The goal of this excersice is to prove that if $p neq 2$ $||.||_p$ is not a norm induced by an inner product (i.e $L^2$ is the only hilbert space among all $L^p$ spaces. )
To do so, study when the parallelogram law holds. Hint: consider the functions
$$f(t)=frac{1}{2}-t;
g(t)=
begin{matrix}
frac{1}{2} -t & text{ if } 0 leq t leq frac{1}{2}
\ t-frac{1}{2} & text{ if } frac{1}{2} <t leq 1 end{matrix} $$
Parallelogram Law $M=C([0,1])$
Let $(M,<.,.>)$ is an inner product space, where induced norm
$$|| .|| =sqrt{<.,.>} $$
then
$$forall x,y in M ; || f+g||^2+||f-g ||^2= 2(||f ||^2+||g||^2) $$
Using u-sub $u=(1/2 -t)$ and $frac{du}{dt}=-1 $ so $dt=-du$
$$ begin{aligned}
||f|| &= int^1_0 (|f(t)|^p dt)^{1/p}= int^1_0 (|1/2 -t|^p dt)^{1/p}
\ &=int^{u(1)}_{u(0)} (u^p -du)^{1/p}
\& =(int^{-1/2}_{1/2} u^p du)^{1/p}
\ &= left [ left (frac{u^{p+1}}{p+1} right )^{1/p} right]^{-1/2}_{1/2}
\&= left [frac{u^{1+1/p}}{(p+1)^{1/p}} right]^{-1/2}_{1/2}
\&=frac{(-1/2)^{1+1/p} -(1/2)^{1+1/p}}{(p+1)^{1/p}}
end{aligned}$$
working on $||f-g ||,||f+g||,||g||$
Guessing big picture that the parallelogram law only works when $p=2$ so it is only induced norm when $p=2$
real-analysis functional-analysis
edited Jan 31 at 19:06
Umberto P.
40.3k13370
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
$endgroup$
add a comment |
$begingroup$
Let the space $C([0,1])$ and consider the norm $forall p in mathbb{N}$
$$forall f in C([0,1]),
||f ||_{L^P}= left ( int^1_0 |f(t)|^P dt right )^{frac{1}{p}} $$
knowing that when $p=2$, this norm is the norm induced by the innder product
$$forall f,g in C([0,1]), <f,g>=int^1_0 f(t)overline{g(t)} dx $$
The goal of this excersice is to prove that if $p neq 2$ $||.||_p$ is not a norm induced by an inner product (i.e $L^2$ is the only hilbert space among all $L^p$ spaces. )
To do so, study when the parallelogram law holds. Hint: consider the functions
$$f(t)=frac{1}{2}-t;
g(t)=
begin{matrix}
frac{1}{2} -t & text{ if } 0 leq t leq frac{1}{2}
\ t-frac{1}{2} & text{ if } frac{1}{2} <t leq 1 end{matrix} $$
Parallelogram Law $M=C([0,1])$
Let $(M,<.,.>)$ is an inner product space, where induced norm
$$|| .|| =sqrt{<.,.>} $$
then
$$forall x,y in M ; || f+g||^2+||f-g ||^2= 2(||f ||^2+||g||^2) $$
Using u-sub $u=(1/2 -t)$ and $frac{du}{dt}=-1 $ so $dt=-du$
$$ begin{aligned}
||f|| &= int^1_0 (|f(t)|^p dt)^{1/p}= int^1_0 (|1/2 -t|^p dt)^{1/p}
\ &=int^{u(1)}_{u(0)} (u^p -du)^{1/p}
\& =(int^{-1/2}_{1/2} u^p du)^{1/p}
\ &= left [ left (frac{u^{p+1}}{p+1} right )^{1/p} right]^{-1/2}_{1/2}
\&= left [frac{u^{1+1/p}}{(p+1)^{1/p}} right]^{-1/2}_{1/2}
\&=frac{(-1/2)^{1+1/p} -(1/2)^{1+1/p}}{(p+1)^{1/p}}
end{aligned}$$
working on $||f-g ||,||f+g||,||g||$
Guessing big picture that the parallelogram law only works when $p=2$ so it is only induced norm when $p=2$
real-analysis functional-analysis
edited Jan 31 at 19:06
Umberto P.
40.3k13370
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
$endgroup$
$begingroup$
When $p=2$ inner product does induce the norm, and when $pneq 2$ parallelogram fails. If norms were induced by inner product parallelogram law "shouldn't" fail, so it cannot possibly be induced by inner products
$endgroup$
– user160738
Nov 8 '16 at 19:49
add a comment |
$begingroup$
Let the space $C([0,1])$ and consider the norm $forall p in mathbb{N}$
$$forall f in C([0,1]),
||f ||_{L^P}= left ( int^1_0 |f(t)|^P dt right )^{frac{1}{p}} $$
knowing that when $p=2$, this norm is the norm induced by the innder product
$$forall f,g in C([0,1]), <f,g>=int^1_0 f(t)overline{g(t)} dx $$
The goal of this excersice is to prove that if $p neq 2$ $||.||_p$ is not a norm induced by an inner product (i.e $L^2$ is the only hilbert space among all $L^p$ spaces. )
To do so, study when the parallelogram law holds. Hint: consider the functions
$$f(t)=frac{1}{2}-t;
g(t)=
begin{matrix}
frac{1}{2} -t & text{ if } 0 leq t leq frac{1}{2}
\ t-frac{1}{2} & text{ if } frac{1}{2} <t leq 1 end{matrix} $$
Parallelogram Law $M=C([0,1])$
Let $(M,<.,.>)$ is an inner product space, where induced norm
$$|| .|| =sqrt{<.,.>} $$
then
$$forall x,y in M ; || f+g||^2+||f-g ||^2= 2(||f ||^2+||g||^2) $$
Using u-sub $u=(1/2 -t)$ and $frac{du}{dt}=-1 $ so $dt=-du$
$$ begin{aligned}
||f|| &= int^1_0 (|f(t)|^p dt)^{1/p}= int^1_0 (|1/2 -t|^p dt)^{1/p}
\ &=int^{u(1)}_{u(0)} (u^p -du)^{1/p}
\& =(int^{-1/2}_{1/2} u^p du)^{1/p}
\ &= left [ left (frac{u^{p+1}}{p+1} right )^{1/p} right]^{-1/2}_{1/2}
\&= left [frac{u^{1+1/p}}{(p+1)^{1/p}} right]^{-1/2}_{1/2}
\&=frac{(-1/2)^{1+1/p} -(1/2)^{1+1/p}}{(p+1)^{1/p}}
end{aligned}$$
working on $||f-g ||,||f+g||,||g||$
Guessing big picture that the parallelogram law only works when $p=2$ so it is only induced norm when $p=2$
real-analysis functional-analysis
edited Jan 31 at 19:06
Umberto P.
40.3k13370
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
$endgroup$
Let the space $C([0,1])$ and consider the norm $forall p in mathbb{N}$
$$forall f in C([0,1]),
||f ||_{L^P}= left ( int^1_0 |f(t)|^P dt right )^{frac{1}{p}} $$
knowing that when $p=2$, this norm is the norm induced by the innder product
$$forall f,g in C([0,1]), <f,g>=int^1_0 f(t)overline{g(t)} dx $$
The goal of this excersice is to prove that if $p neq 2$ $||.||_p$ is not a norm induced by an inner product (i.e $L^2$ is the only hilbert space among all $L^p$ spaces. )
To do so, study when the parallelogram law holds. Hint: consider the functions
$$f(t)=frac{1}{2}-t;
g(t)=
begin{matrix}
frac{1}{2} -t & text{ if } 0 leq t leq frac{1}{2}
\ t-frac{1}{2} & text{ if } frac{1}{2} <t leq 1 end{matrix} $$
Parallelogram Law $M=C([0,1])$
Let $(M,<.,.>)$ is an inner product space, where induced norm
$$|| .|| =sqrt{<.,.>} $$
then
$$forall x,y in M ; || f+g||^2+||f-g ||^2= 2(||f ||^2+||g||^2) $$
Using u-sub $u=(1/2 -t)$ and $frac{du}{dt}=-1 $ so $dt=-du$
$$ begin{aligned}
||f|| &= int^1_0 (|f(t)|^p dt)^{1/p}= int^1_0 (|1/2 -t|^p dt)^{1/p}
\ &=int^{u(1)}_{u(0)} (u^p -du)^{1/p}
\& =(int^{-1/2}_{1/2} u^p du)^{1/p}
\ &= left [ left (frac{u^{p+1}}{p+1} right )^{1/p} right]^{-1/2}_{1/2}
\&= left [frac{u^{1+1/p}}{(p+1)^{1/p}} right]^{-1/2}_{1/2}
\&=frac{(-1/2)^{1+1/p} -(1/2)^{1+1/p}}{(p+1)^{1/p}}
end{aligned}$$
working on $||f-g ||,||f+g||,||g||$
Guessing big picture that the parallelogram law only works when $p=2$ so it is only induced norm when $p=2$
real-analysis functional-analysis
real-analysis functional-analysis
edited Jan 31 at 19:06
Umberto P.
40.3k13370
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
edited Jan 31 at 19:06
Umberto P.
40.3k13370
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
edited Jan 31 at 19:06
Umberto P.
40.3k13370
edited Jan 31 at 19:06
Umberto P.
40.3k13370
edited Jan 31 at 19:06
Umberto P.
40.3k13370
40.3k13370
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
asked Nov 8 '16 at 19:45
Tiger BloodTiger Blood
827726
827726
$begingroup$
When $p=2$ inner product does induce the norm, and when $pneq 2$ parallelogram fails. If norms were induced by inner product parallelogram law "shouldn't" fail, so it cannot possibly be induced by inner products
$endgroup$
– user160738
Nov 8 '16 at 19:49
add a comment |
$begingroup$
When $p=2$ inner product does induce the norm, and when $pneq 2$ parallelogram fails. If norms were induced by inner product parallelogram law "shouldn't" fail, so it cannot possibly be induced by inner products
$endgroup$
– user160738
Nov 8 '16 at 19:49
$begingroup$
When $p=2$ inner product does induce the norm, and when $pneq 2$ parallelogram fails. If norms were induced by inner product parallelogram law "shouldn't" fail, so it cannot possibly be induced by inner products
$endgroup$
– user160738
Nov 8 '16 at 19:49
$begingroup$
When $p=2$ inner product does induce the norm, and when $pneq 2$ parallelogram fails. If norms were induced by inner product parallelogram law "shouldn't" fail, so it cannot possibly be induced by inner products
$endgroup$
– user160738
Nov 8 '16 at 19:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall that Hilbert spaces are self-dual via the Riesz Representation theorem. But then as $1<p<infty$ we know that ${L^p}^* = L^q$ where $q$ is the Holder conjugate of $p$. And when $pne q$ these are not isomorphic.
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
$endgroup$
$begingroup$
How do you prove that $L^p$ and $L^q$ are not isomorphic for $p ne q$?
$endgroup$
– gerw
Nov 8 '16 at 19:59
1
$begingroup$
This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one..
$endgroup$
– gerw
Nov 8 '16 at 20:05
add a comment |
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1 Answer
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$begingroup$
Recall that Hilbert spaces are self-dual via the Riesz Representation theorem. But then as $1<p<infty$ we know that ${L^p}^* = L^q$ where $q$ is the Holder conjugate of $p$. And when $pne q$ these are not isomorphic.
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
$endgroup$
$begingroup$
How do you prove that $L^p$ and $L^q$ are not isomorphic for $p ne q$?
$endgroup$
– gerw
Nov 8 '16 at 19:59
1
$begingroup$
This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one..
$endgroup$
– gerw
Nov 8 '16 at 20:05
add a comment |
$begingroup$
Recall that Hilbert spaces are self-dual via the Riesz Representation theorem. But then as $1<p<infty$ we know that ${L^p}^* = L^q$ where $q$ is the Holder conjugate of $p$. And when $pne q$ these are not isomorphic.
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
$endgroup$
$begingroup$
How do you prove that $L^p$ and $L^q$ are not isomorphic for $p ne q$?
$endgroup$
– gerw
Nov 8 '16 at 19:59
1
$begingroup$
This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one..
$endgroup$
– gerw
Nov 8 '16 at 20:05
add a comment |
$begingroup$
Recall that Hilbert spaces are self-dual via the Riesz Representation theorem. But then as $1<p<infty$ we know that ${L^p}^* = L^q$ where $q$ is the Holder conjugate of $p$. And when $pne q$ these are not isomorphic.
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
$endgroup$
Recall that Hilbert spaces are self-dual via the Riesz Representation theorem. But then as $1<p<infty$ we know that ${L^p}^* = L^q$ where $q$ is the Holder conjugate of $p$. And when $pne q$ these are not isomorphic.
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
edited Nov 8 '16 at 20:01
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
answered Nov 8 '16 at 19:58
Adam HughesAdam Hughes
32.3k83770
32.3k83770
$begingroup$
How do you prove that $L^p$ and $L^q$ are not isomorphic for $p ne q$?
$endgroup$
– gerw
Nov 8 '16 at 19:59
1
$begingroup$
This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one..
$endgroup$
– gerw
Nov 8 '16 at 20:05
add a comment |
$begingroup$
How do you prove that $L^p$ and $L^q$ are not isomorphic for $p ne q$?
$endgroup$
– gerw
Nov 8 '16 at 19:59
1
$begingroup$
This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one..
$endgroup$
– gerw
Nov 8 '16 at 20:05
$begingroup$
How do you prove that $L^p$ and $L^q$ are not isomorphic for $p ne q$?
$endgroup$
– gerw
Nov 8 '16 at 19:59
$begingroup$
How do you prove that $L^p$ and $L^q$ are not isomorphic for $p ne q$?
$endgroup$
– gerw
Nov 8 '16 at 19:59
1
1
$begingroup$
This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one..
$endgroup$
– gerw
Nov 8 '16 at 20:05
$begingroup$
This does only show that the identity is not an isomorphism between $L^p$ and $L^q$. However, there might be another one..
$endgroup$
– gerw
Nov 8 '16 at 20:05
add a comment |
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$begingroup$
When $p=2$ inner product does induce the norm, and when $pneq 2$ parallelogram fails. If norms were induced by inner product parallelogram law "shouldn't" fail, so it cannot possibly be induced by inner products
$endgroup$
– user160738
Nov 8 '16 at 19:49