$L^p $ Spaces
Let's consider $L^p(RR)$ spaces with $(1<=p< oo)$.
Let's prove that only for $ p=2 $ the norm is originated by a scalar product.
Let's prove that only for $ p=2 $ the norm is originated by a scalar product.
Risposte
Ok! You're right! I used the thesis for proving them!
A bad mistake.
A bad mistake.
I think there is a mistake in your proof: why $\int fg dx$ is a scalar product in $L^p$? This forces $p=2$, and the proof is essentially Camillo's proof.
I don't know how to insert a diagram ...
However:
for all $p\in[1,\infty)$ we have the following duality
$S : (L^p)$*$->L^{(1-1/p)^{-1}}$
$F->f:F(h)=\intfhdx, \forall h\inL^p$
I suppose now that $L^p$ is an Hilbert space. Thus there is the Riesz
represantion $R$
$R : (L^p)$*$->L^p$
$F->g:F(h)=\intgh, \forall h\inL^p$
I observe that $R(F)=S(F),\forall F\inL^p$*. Infact, the condition
$\intgh=\intfh,\forall h\in L^p$ leads to $f=g$ (for example we can consider
only test functions $h$ and apply the well-knowed theorem on test functions).
So, I can prove that $SR^{-1}=I$ and $RS^{-1}=I$. This proves that the
diagram commutes with $I$. This forces $p=(1-1/p)^{-1}$.
Now I'm quite sure that it is correct ... but maybe I am wrong!
However:
for all $p\in[1,\infty)$ we have the following duality
$S : (L^p)$*$->L^{(1-1/p)^{-1}}$
$F->f:F(h)=\intfhdx, \forall h\inL^p$
I suppose now that $L^p$ is an Hilbert space. Thus there is the Riesz
represantion $R$
$R : (L^p)$*$->L^p$
$F->g:F(h)=\intgh, \forall h\inL^p$
I observe that $R(F)=S(F),\forall F\inL^p$*. Infact, the condition
$\intgh=\intfh,\forall h\in L^p$ leads to $f=g$ (for example we can consider
only test functions $h$ and apply the well-knowed theorem on test functions).
So, I can prove that $SR^{-1}=I$ and $RS^{-1}=I$. This proves that the
diagram commutes with $I$. This forces $p=(1-1/p)^{-1}$.
Now I'm quite sure that it is correct ... but maybe I am wrong!
I don't know if it is correct or not... but why the map that closes the diagram should be the identity??
It's not easy!!
I'd like to know why (and where!) my proof is not correct ...
I'd like to know why (and where!) my proof is not correct ...
I indicate herebelow the standard solution I had in mind .
Let's be $f(x) = chi_([0,1]) (x) ; g(x) = chi_([1,2]) (x)$, where $chi_([a,b])(x)]$ is the characteristic function of the set $[a,b]$.
We get :
$||f+g||_p = ||f-g||_p =(int_0^2 1^p*dx)^(1/p)=2^(1/p) $ .
and also
$||f ||_p = ||g||_p = 1 $ .
Consequently :
a) $ ||f+g||_p^2 +||f-g||_p^2 = 2^(2/p)+2^(2/p) = 2*2^(2/p) $ .
b) $ 2(||f||_p^2 +||g||_p^2 ) = 2*2 $.
The two expressions a),b) are equal iff $2^(2/p) = 2 $ that is $ p=2 $ .
For $ pne 2 $ the equality of parallelogram is phalse.
For $ p=2 $ we already know that
$< f,g> =int_R f(x)*g(x) $ is a scalar product which induces the $L^2(R) $ norm
.
Let's be $f(x) = chi_([0,1]) (x) ; g(x) = chi_([1,2]) (x)$, where $chi_([a,b])(x)]$ is the characteristic function of the set $[a,b]$.
We get :
$||f+g||_p = ||f-g||_p =(int_0^2 1^p*dx)^(1/p)=2^(1/p) $ .
and also
$||f ||_p = ||g||_p = 1 $ .
Consequently :
a) $ ||f+g||_p^2 +||f-g||_p^2 = 2^(2/p)+2^(2/p) = 2*2^(2/p) $ .
b) $ 2(||f||_p^2 +||g||_p^2 ) = 2*2 $.
The two expressions a),b) are equal iff $2^(2/p) = 2 $ that is $ p=2 $ .
For $ pne 2 $ the equality of parallelogram is phalse.
For $ p=2 $ we already know that
$< f,g> =int_R f(x)*g(x) $ is a scalar product which induces the $L^2(R) $ norm
.
"ubermensch":
the linear isometric isomorphism
between $L^{1-1/p}$ e $L^p$ is realized by the identity map and this certainly forces $p=1-1/p$.
Is here the "error"?
I am not sure that the uber's proof works; I certainly consider correct this proof only in the case of finite dimensional vector spaces. But $L^p$ are infinite dimensional vector spaces....
I will leave to Luca the task to judge if what you wrote is correct or not 
Later I will write the solution I had in mind, based on the parallelogram rule .
Later I will write the solution I had in mind, based on the parallelogram rule .
You don't disturb ... and my english is probably worse than your.
Howewer, I thought that the linear isometric isomorphism $F: (L^p)$*$->L^{1-1/p}$
is defined by $F(f)=g$ such that $f(h)=\int ghdx, \forall h\inL^p$.
On the other hand, the Riesz representation is the same than $F$(but from $(L^p)$* to $L^p$).
Thus, I hope (using a commutative diagram!) that the linear isometric isomorphism
between $L^{1-1/p}$ e $L^p$ is realized by the identity map and this certainly forces $p=1-1/p$.
In conclusion, I think that $L^p$ linearly isometric to $L^q$ does not imply $p=q$, but this is true
in our particular case.
Quante cavolate ho detto?
Howewer, I thought that the linear isometric isomorphism $F: (L^p)$*$->L^{1-1/p}$
is defined by $F(f)=g$ such that $f(h)=\int ghdx, \forall h\inL^p$.
On the other hand, the Riesz representation is the same than $F$(but from $(L^p)$* to $L^p$).
Thus, I hope (using a commutative diagram!) that the linear isometric isomorphism
between $L^{1-1/p}$ e $L^p$ is realized by the identity map and this certainly forces $p=1-1/p$.
In conclusion, I think that $L^p$ linearly isometric to $L^q$ does not imply $p=q$, but this is true
in our particular case.
Quante cavolate ho detto?
But, could one use parallelogram's rule? Is it possible or not? (Sorry for my bad English, if I disturb, I'll go away...)
ooops No!! I'm not sure ... I'didn't find the answer in my book of functional analysis; thus it is difficult...
or trivial
or trivial
Are you sure that if $L^p$ and $L^q$ are linearly isomorphic then $p=q$? You can prove it?
The dual of a real Hilbert space is linearly hysomorphic to itself and thus... $p=2$.
This argument don't work for $p=\infty$. But $L^{\infty}$ is not uniformly convex,
while each Hilbert space is it. (another solution: a Hilbert space is reflexive and
$L^{\infty}$ is not reflexive (because it is dual of a separable space and it isn't
separable!))
p.s. the dual of a complex hilbert space is its conjugate!
This argument don't work for $p=\infty$. But $L^{\infty}$ is not uniformly convex,
while each Hilbert space is it. (another solution: a Hilbert space is reflexive and
$L^{\infty}$ is not reflexive (because it is dual of a separable space and it isn't
separable!))
p.s. the dual of a complex hilbert space is its conjugate!
Accedi a tutti gli appunti
Tutor AI: studia meglio e in meno tempo