Show that this is an inner product
$begingroup$
Let's define
$$(f,g)=int_{mathbb{R}} frac{f(x)bar{g}(x)}{1+x^2}dx$$
$forall f,gin X={h:mathbb{R}rightarrowmathbb{C}:$ $h$ is Lebesgue-measurable and bounded over $mathbb{R}$}
I have to show that $(cdot,cdot)$ is an inner product over $X$ (I've already shown that $X$ is a vector space over $mathbb{C}$).
It is obvious that, $forall f,g,hin X$, $(f+g,h)=(f,h)+(g,h)$ and, $forall lambdainmathbb{C}$, that $(lambda f,g)=lambda(f,g)$.
It's also clear that $(f,g)=overline{(g,f)}$, so we obtain the skew-linearity in the second argument. Now I should show that $(f,f)geq 0$ $forall fin X$ (this is trivial) and that $(f,f)=0 Rightarrow f=0$, but this seems to be false, because any $fin X$ which is almost everywhere $0$ over $mathbb{R}$ would be such that $(f,f)=0$, even if $f$ is not identically $0$. Did I do some mistakes in this argument?
hilbert-spaces lebesgue-integral lebesgue-measure inner-product-space almost-everywhere
asked Jan 16 at 16:29
LukathLukath
717
$endgroup$
add a comment |
$begingroup$
Let's define
$$(f,g)=int_{mathbb{R}} frac{f(x)bar{g}(x)}{1+x^2}dx$$
$forall f,gin X={h:mathbb{R}rightarrowmathbb{C}:$ $h$ is Lebesgue-measurable and bounded over $mathbb{R}$}
I have to show that $(cdot,cdot)$ is an inner product over $X$ (I've already shown that $X$ is a vector space over $mathbb{C}$).
It is obvious that, $forall f,g,hin X$, $(f+g,h)=(f,h)+(g,h)$ and, $forall lambdainmathbb{C}$, that $(lambda f,g)=lambda(f,g)$.
It's also clear that $(f,g)=overline{(g,f)}$, so we obtain the skew-linearity in the second argument. Now I should show that $(f,f)geq 0$ $forall fin X$ (this is trivial) and that $(f,f)=0 Rightarrow f=0$, but this seems to be false, because any $fin X$ which is almost everywhere $0$ over $mathbb{R}$ would be such that $(f,f)=0$, even if $f$ is not identically $0$. Did I do some mistakes in this argument?
hilbert-spaces lebesgue-integral lebesgue-measure inner-product-space almost-everywhere
asked Jan 16 at 16:29
LukathLukath
717
$endgroup$
1
$begingroup$
You'll probably get the answer to your question if you tell us precisely on which space your inner product is defined.
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:33
$begingroup$
Ok sorry, I've edited the question, I hope it's clear now! @mathcounterexamples.net
$endgroup$
– Lukath
Jan 16 at 16:45
add a comment |
$begingroup$
Let's define
$$(f,g)=int_{mathbb{R}} frac{f(x)bar{g}(x)}{1+x^2}dx$$
$forall f,gin X={h:mathbb{R}rightarrowmathbb{C}:$ $h$ is Lebesgue-measurable and bounded over $mathbb{R}$}
I have to show that $(cdot,cdot)$ is an inner product over $X$ (I've already shown that $X$ is a vector space over $mathbb{C}$).
It is obvious that, $forall f,g,hin X$, $(f+g,h)=(f,h)+(g,h)$ and, $forall lambdainmathbb{C}$, that $(lambda f,g)=lambda(f,g)$.
It's also clear that $(f,g)=overline{(g,f)}$, so we obtain the skew-linearity in the second argument. Now I should show that $(f,f)geq 0$ $forall fin X$ (this is trivial) and that $(f,f)=0 Rightarrow f=0$, but this seems to be false, because any $fin X$ which is almost everywhere $0$ over $mathbb{R}$ would be such that $(f,f)=0$, even if $f$ is not identically $0$. Did I do some mistakes in this argument?
hilbert-spaces lebesgue-integral lebesgue-measure inner-product-space almost-everywhere
asked Jan 16 at 16:29
LukathLukath
717
$endgroup$
Let's define
$$(f,g)=int_{mathbb{R}} frac{f(x)bar{g}(x)}{1+x^2}dx$$
$forall f,gin X={h:mathbb{R}rightarrowmathbb{C}:$ $h$ is Lebesgue-measurable and bounded over $mathbb{R}$}
I have to show that $(cdot,cdot)$ is an inner product over $X$ (I've already shown that $X$ is a vector space over $mathbb{C}$).
It is obvious that, $forall f,g,hin X$, $(f+g,h)=(f,h)+(g,h)$ and, $forall lambdainmathbb{C}$, that $(lambda f,g)=lambda(f,g)$.
It's also clear that $(f,g)=overline{(g,f)}$, so we obtain the skew-linearity in the second argument. Now I should show that $(f,f)geq 0$ $forall fin X$ (this is trivial) and that $(f,f)=0 Rightarrow f=0$, but this seems to be false, because any $fin X$ which is almost everywhere $0$ over $mathbb{R}$ would be such that $(f,f)=0$, even if $f$ is not identically $0$. Did I do some mistakes in this argument?
hilbert-spaces lebesgue-integral lebesgue-measure inner-product-space almost-everywhere
hilbert-spaces lebesgue-integral lebesgue-measure inner-product-space almost-everywhere
asked Jan 16 at 16:29
LukathLukath
717
asked Jan 16 at 16:29
LukathLukath
717
edited Jan 16 at 16:44
Lukath
asked Jan 16 at 16:29
LukathLukath
717
asked Jan 16 at 16:29
LukathLukath
717
asked Jan 16 at 16:29
LukathLukath
717
717
1
$begingroup$
You'll probably get the answer to your question if you tell us precisely on which space your inner product is defined.
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:33
$begingroup$
Ok sorry, I've edited the question, I hope it's clear now! @mathcounterexamples.net
$endgroup$
– Lukath
Jan 16 at 16:45
add a comment |
1
$begingroup$
You'll probably get the answer to your question if you tell us precisely on which space your inner product is defined.
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:33
$begingroup$
Ok sorry, I've edited the question, I hope it's clear now! @mathcounterexamples.net
$endgroup$
– Lukath
Jan 16 at 16:45
1
1
$begingroup$
You'll probably get the answer to your question if you tell us precisely on which space your inner product is defined.
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:33
$begingroup$
You'll probably get the answer to your question if you tell us precisely on which space your inner product is defined.
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:33
$begingroup$
Ok sorry, I've edited the question, I hope it's clear now! @mathcounterexamples.net
$endgroup$
– Lukath
Jan 16 at 16:45
$begingroup$
Ok sorry, I've edited the question, I hope it's clear now! @mathcounterexamples.net
$endgroup$
– Lukath
Jan 16 at 16:45
add a comment |
1 Answer
1
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oldest
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$begingroup$
You are right that $(f, f)=0$ doesn't implies that $fequiv 0$ as a function. Nevertheless, there's a sense in which $(cdot, cdot)$ can be an inner product in some space.
Consider the space $mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)$ consisting of functions $f:Bbb Rto Bbb C$ such that
$$
int_{Bbb R} frac {|f|^2}{1+x^2}dx < infty.
$$
We can then quotient out by the relation $gsim f$ iff $f=g$ almost everywhere. Then the elements of
$$
L^2left(Bbb R,frac 1{1+x^2},dxright):= mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)Big/ sim
$$
would consist of equivalent classes of functions instead.
Now, for $fin L^2left(Bbb R,frac 1{1+x^2},dxright)$ if $(f, f)=0$ then, as you've shown, we would have $f=0$ almost everywhere. This means that $f$ belong to the equivalent class of the constant function $0$, so $f$ is the zero vector in $ L^2left(Bbb R,frac 1{1+x^2},dxright)$.
Note that this process of quotient out by $sim$ as I mentioned is the usual one used in the construction of $L^p(Omega)$ spaces in functional analysis.
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
$endgroup$
add a comment |
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$begingroup$
You are right that $(f, f)=0$ doesn't implies that $fequiv 0$ as a function. Nevertheless, there's a sense in which $(cdot, cdot)$ can be an inner product in some space.
Consider the space $mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)$ consisting of functions $f:Bbb Rto Bbb C$ such that
$$
int_{Bbb R} frac {|f|^2}{1+x^2}dx < infty.
$$
We can then quotient out by the relation $gsim f$ iff $f=g$ almost everywhere. Then the elements of
$$
L^2left(Bbb R,frac 1{1+x^2},dxright):= mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)Big/ sim
$$
would consist of equivalent classes of functions instead.
Now, for $fin L^2left(Bbb R,frac 1{1+x^2},dxright)$ if $(f, f)=0$ then, as you've shown, we would have $f=0$ almost everywhere. This means that $f$ belong to the equivalent class of the constant function $0$, so $f$ is the zero vector in $ L^2left(Bbb R,frac 1{1+x^2},dxright)$.
Note that this process of quotient out by $sim$ as I mentioned is the usual one used in the construction of $L^p(Omega)$ spaces in functional analysis.
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
$endgroup$
add a comment |
$begingroup$
You are right that $(f, f)=0$ doesn't implies that $fequiv 0$ as a function. Nevertheless, there's a sense in which $(cdot, cdot)$ can be an inner product in some space.
Consider the space $mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)$ consisting of functions $f:Bbb Rto Bbb C$ such that
$$
int_{Bbb R} frac {|f|^2}{1+x^2}dx < infty.
$$
We can then quotient out by the relation $gsim f$ iff $f=g$ almost everywhere. Then the elements of
$$
L^2left(Bbb R,frac 1{1+x^2},dxright):= mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)Big/ sim
$$
would consist of equivalent classes of functions instead.
Now, for $fin L^2left(Bbb R,frac 1{1+x^2},dxright)$ if $(f, f)=0$ then, as you've shown, we would have $f=0$ almost everywhere. This means that $f$ belong to the equivalent class of the constant function $0$, so $f$ is the zero vector in $ L^2left(Bbb R,frac 1{1+x^2},dxright)$.
Note that this process of quotient out by $sim$ as I mentioned is the usual one used in the construction of $L^p(Omega)$ spaces in functional analysis.
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
$endgroup$
add a comment |
$begingroup$
You are right that $(f, f)=0$ doesn't implies that $fequiv 0$ as a function. Nevertheless, there's a sense in which $(cdot, cdot)$ can be an inner product in some space.
Consider the space $mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)$ consisting of functions $f:Bbb Rto Bbb C$ such that
$$
int_{Bbb R} frac {|f|^2}{1+x^2}dx < infty.
$$
We can then quotient out by the relation $gsim f$ iff $f=g$ almost everywhere. Then the elements of
$$
L^2left(Bbb R,frac 1{1+x^2},dxright):= mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)Big/ sim
$$
would consist of equivalent classes of functions instead.
Now, for $fin L^2left(Bbb R,frac 1{1+x^2},dxright)$ if $(f, f)=0$ then, as you've shown, we would have $f=0$ almost everywhere. This means that $f$ belong to the equivalent class of the constant function $0$, so $f$ is the zero vector in $ L^2left(Bbb R,frac 1{1+x^2},dxright)$.
Note that this process of quotient out by $sim$ as I mentioned is the usual one used in the construction of $L^p(Omega)$ spaces in functional analysis.
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
$endgroup$
You are right that $(f, f)=0$ doesn't implies that $fequiv 0$ as a function. Nevertheless, there's a sense in which $(cdot, cdot)$ can be an inner product in some space.
Consider the space $mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)$ consisting of functions $f:Bbb Rto Bbb C$ such that
$$
int_{Bbb R} frac {|f|^2}{1+x^2}dx < infty.
$$
We can then quotient out by the relation $gsim f$ iff $f=g$ almost everywhere. Then the elements of
$$
L^2left(Bbb R,frac 1{1+x^2},dxright):= mathcal L^2left(Bbb R,frac 1{1+x^2},dxright)Big/ sim
$$
would consist of equivalent classes of functions instead.
Now, for $fin L^2left(Bbb R,frac 1{1+x^2},dxright)$ if $(f, f)=0$ then, as you've shown, we would have $f=0$ almost everywhere. This means that $f$ belong to the equivalent class of the constant function $0$, so $f$ is the zero vector in $ L^2left(Bbb R,frac 1{1+x^2},dxright)$.
Note that this process of quotient out by $sim$ as I mentioned is the usual one used in the construction of $L^p(Omega)$ spaces in functional analysis.
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
answered Jan 16 at 16:45
BigbearZzzBigbearZzz
9,00521652
9,00521652
add a comment |
add a comment |
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$begingroup$
You'll probably get the answer to your question if you tell us precisely on which space your inner product is defined.
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:33
$begingroup$
Ok sorry, I've edited the question, I hope it's clear now! @mathcounterexamples.net
$endgroup$
– Lukath
Jan 16 at 16:45