Proof no nonnegative function satisfies 3 integrals (3 $L^2$ norms derived from the inner product on $C[0,1]$)Inner product and norms for random vectorsNorms on inner product space over $mathbbR$Real inner product from a complex onePossible ways to induce norm from inner productDoes the inner product $langle cdot, cdot rangle$ induce any other norms other than the 2 norm?Calculate the norm of a polynomial knowing inner productEquivalence of Norms and Norms from an Inner productInner Product and Norms of vectorsDoes a norm induce an inner product uniquely in the sense that $|x|^2 = langle x,xrangle?$Can we define an inner product in terms of the norm induced by it?
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Proof no nonnegative function satisfies 3 integrals (3 $L^2$ norms derived from the inner product on $C[0,1]$)
Inner product and norms for random vectorsNorms on inner product space over $mathbbR$Real inner product from a complex onePossible ways to induce norm from inner productDoes the inner product $langle cdot, cdot rangle$ induce any other norms other than the 2 norm?Calculate the norm of a polynomial knowing inner productEquivalence of Norms and Norms from an Inner productInner Product and Norms of vectorsDoes a norm induce an inner product uniquely in the sense that $|x|^2 = langle x,xrangle?$Can we define an inner product in terms of the norm induced by it?
$begingroup$
$V$ is the real inner product space $C_mathbbR[0,1]$ and $a in [0,1]$. Prove there is no nonnegative function $f in V$ such that
$$ int_0^1 f(x) , dx = 1,$$
$$ int_0^1 xf(x) , dx = a, $$
$$ mathrmand int_0^1 x^2 f(x) , dx = a^2 $$
We have that the norm derived from the $L^2$ inner product on $C[0,1]$ is the $L^2$ norm
$$ langle g, g rangle = Vert g Vert^2 = int _0^1 |g(t)|^2 , dt $$
We also have that the Cauchy-Schwartz inequality states that:
$$ langle u,v rangle leq Vert u Vert Vert v Vert $$
(With equality if and only if $u$ and $v$ are scalar multiples.) Here, for an inner product of something with itself, C.-S. becomes:
$$ langle g,g rangle = Vert g Vert^2 $$
So given the above 3 integrals, we have
$$g_1 = f(x)$$
$$g_2 = xf(x)$$
$$g_3 = x^2f(x)$$
And accordingly,
$$ Vert g_1 Vert^2 = 1, , g_1 = sqrtf(x)$$
$$ Vert g_2 Vert^2 = a, , g_2 = sqrtxf(x)$$
$$ Vert g_3 Vert^2 = a^2, , g_3 = sqrtx^2f(x)$$
However, after several attempts, I can't seem to finish the proof along this line of thought. Perhaps I'm squaring or square-rooting something incorrectly? I suspect that when we square $a$ from the second integral and try to equate things with $a^2$ from the third, something goes wrong with Cauchy-Schwartz. But how to bring it home?
linear-algebra functional-analysis inner-product-space
asked Mar 22 at 3:55
user636164user636164
475
$endgroup$
add a comment |
$begingroup$
$V$ is the real inner product space $C_mathbbR[0,1]$ and $a in [0,1]$. Prove there is no nonnegative function $f in V$ such that
$$ int_0^1 f(x) , dx = 1,$$
$$ int_0^1 xf(x) , dx = a, $$
$$ mathrmand int_0^1 x^2 f(x) , dx = a^2 $$
We have that the norm derived from the $L^2$ inner product on $C[0,1]$ is the $L^2$ norm
$$ langle g, g rangle = Vert g Vert^2 = int _0^1 |g(t)|^2 , dt $$
We also have that the Cauchy-Schwartz inequality states that:
$$ langle u,v rangle leq Vert u Vert Vert v Vert $$
(With equality if and only if $u$ and $v$ are scalar multiples.) Here, for an inner product of something with itself, C.-S. becomes:
$$ langle g,g rangle = Vert g Vert^2 $$
So given the above 3 integrals, we have
$$g_1 = f(x)$$
$$g_2 = xf(x)$$
$$g_3 = x^2f(x)$$
And accordingly,
$$ Vert g_1 Vert^2 = 1, , g_1 = sqrtf(x)$$
$$ Vert g_2 Vert^2 = a, , g_2 = sqrtxf(x)$$
$$ Vert g_3 Vert^2 = a^2, , g_3 = sqrtx^2f(x)$$
However, after several attempts, I can't seem to finish the proof along this line of thought. Perhaps I'm squaring or square-rooting something incorrectly? I suspect that when we square $a$ from the second integral and try to equate things with $a^2$ from the third, something goes wrong with Cauchy-Schwartz. But how to bring it home?
linear-algebra functional-analysis inner-product-space
asked Mar 22 at 3:55
user636164user636164
475
$endgroup$
1
$begingroup$
Given the information, $f$ can be seen as a probability density function of a r.v $Xin [0,1]$ such that $Bbb E[X]=a, Bbb E[X^2]=a^2$. Then, $textvar(X)=Bbb E[X^2]-Bbb E[X]^2=0$, which implies $Bbb P(X=a)=1$, but then $X$ cannot have a continuous pdf.
$endgroup$
– Song
Mar 22 at 4:09
add a comment |
$begingroup$
$V$ is the real inner product space $C_mathbbR[0,1]$ and $a in [0,1]$. Prove there is no nonnegative function $f in V$ such that
$$ int_0^1 f(x) , dx = 1,$$
$$ int_0^1 xf(x) , dx = a, $$
$$ mathrmand int_0^1 x^2 f(x) , dx = a^2 $$
We have that the norm derived from the $L^2$ inner product on $C[0,1]$ is the $L^2$ norm
$$ langle g, g rangle = Vert g Vert^2 = int _0^1 |g(t)|^2 , dt $$
We also have that the Cauchy-Schwartz inequality states that:
$$ langle u,v rangle leq Vert u Vert Vert v Vert $$
(With equality if and only if $u$ and $v$ are scalar multiples.) Here, for an inner product of something with itself, C.-S. becomes:
$$ langle g,g rangle = Vert g Vert^2 $$
So given the above 3 integrals, we have
$$g_1 = f(x)$$
$$g_2 = xf(x)$$
$$g_3 = x^2f(x)$$
And accordingly,
$$ Vert g_1 Vert^2 = 1, , g_1 = sqrtf(x)$$
$$ Vert g_2 Vert^2 = a, , g_2 = sqrtxf(x)$$
$$ Vert g_3 Vert^2 = a^2, , g_3 = sqrtx^2f(x)$$
However, after several attempts, I can't seem to finish the proof along this line of thought. Perhaps I'm squaring or square-rooting something incorrectly? I suspect that when we square $a$ from the second integral and try to equate things with $a^2$ from the third, something goes wrong with Cauchy-Schwartz. But how to bring it home?
linear-algebra functional-analysis inner-product-space
asked Mar 22 at 3:55
user636164user636164
475
$endgroup$
$V$ is the real inner product space $C_mathbbR[0,1]$ and $a in [0,1]$. Prove there is no nonnegative function $f in V$ such that
$$ int_0^1 f(x) , dx = 1,$$
$$ int_0^1 xf(x) , dx = a, $$
$$ mathrmand int_0^1 x^2 f(x) , dx = a^2 $$
We have that the norm derived from the $L^2$ inner product on $C[0,1]$ is the $L^2$ norm
$$ langle g, g rangle = Vert g Vert^2 = int _0^1 |g(t)|^2 , dt $$
We also have that the Cauchy-Schwartz inequality states that:
$$ langle u,v rangle leq Vert u Vert Vert v Vert $$
(With equality if and only if $u$ and $v$ are scalar multiples.) Here, for an inner product of something with itself, C.-S. becomes:
$$ langle g,g rangle = Vert g Vert^2 $$
So given the above 3 integrals, we have
$$g_1 = f(x)$$
$$g_2 = xf(x)$$
$$g_3 = x^2f(x)$$
And accordingly,
$$ Vert g_1 Vert^2 = 1, , g_1 = sqrtf(x)$$
$$ Vert g_2 Vert^2 = a, , g_2 = sqrtxf(x)$$
$$ Vert g_3 Vert^2 = a^2, , g_3 = sqrtx^2f(x)$$
However, after several attempts, I can't seem to finish the proof along this line of thought. Perhaps I'm squaring or square-rooting something incorrectly? I suspect that when we square $a$ from the second integral and try to equate things with $a^2$ from the third, something goes wrong with Cauchy-Schwartz. But how to bring it home?
linear-algebra functional-analysis inner-product-space
linear-algebra functional-analysis inner-product-space
asked Mar 22 at 3:55
user636164user636164
475
asked Mar 22 at 3:55
user636164user636164
475
edited Mar 29 at 20:46
user636164
asked Mar 22 at 3:55
user636164user636164
475
asked Mar 22 at 3:55
user636164user636164
475
asked Mar 22 at 3:55
user636164user636164
475
475
1
$begingroup$
Given the information, $f$ can be seen as a probability density function of a r.v $Xin [0,1]$ such that $Bbb E[X]=a, Bbb E[X^2]=a^2$. Then, $textvar(X)=Bbb E[X^2]-Bbb E[X]^2=0$, which implies $Bbb P(X=a)=1$, but then $X$ cannot have a continuous pdf.
$endgroup$
– Song
Mar 22 at 4:09
add a comment |
1
$begingroup$
Given the information, $f$ can be seen as a probability density function of a r.v $Xin [0,1]$ such that $Bbb E[X]=a, Bbb E[X^2]=a^2$. Then, $textvar(X)=Bbb E[X^2]-Bbb E[X]^2=0$, which implies $Bbb P(X=a)=1$, but then $X$ cannot have a continuous pdf.
$endgroup$
– Song
Mar 22 at 4:09
1
1
$begingroup$
Given the information, $f$ can be seen as a probability density function of a r.v $Xin [0,1]$ such that $Bbb E[X]=a, Bbb E[X^2]=a^2$. Then, $textvar(X)=Bbb E[X^2]-Bbb E[X]^2=0$, which implies $Bbb P(X=a)=1$, but then $X$ cannot have a continuous pdf.
$endgroup$
– Song
Mar 22 at 4:09
$begingroup$
Given the information, $f$ can be seen as a probability density function of a r.v $Xin [0,1]$ such that $Bbb E[X]=a, Bbb E[X^2]=a^2$. Then, $textvar(X)=Bbb E[X^2]-Bbb E[X]^2=0$, which implies $Bbb P(X=a)=1$, but then $X$ cannot have a continuous pdf.
$endgroup$
– Song
Mar 22 at 4:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The three integrals imply that $int_0^1(x-a)^2 f(x) dx=0$, which implies that $f(x)equiv 0$. But this does not satisfy the first integral.
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
$endgroup$
add a comment |
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$begingroup$
The three integrals imply that $int_0^1(x-a)^2 f(x) dx=0$, which implies that $f(x)equiv 0$. But this does not satisfy the first integral.
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
$endgroup$
add a comment |
$begingroup$
The three integrals imply that $int_0^1(x-a)^2 f(x) dx=0$, which implies that $f(x)equiv 0$. But this does not satisfy the first integral.
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
$endgroup$
add a comment |
$begingroup$
The three integrals imply that $int_0^1(x-a)^2 f(x) dx=0$, which implies that $f(x)equiv 0$. But this does not satisfy the first integral.
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
$endgroup$
The three integrals imply that $int_0^1(x-a)^2 f(x) dx=0$, which implies that $f(x)equiv 0$. But this does not satisfy the first integral.
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
answered Mar 22 at 4:02
Poon LeviPoon Levi
67639
67639
add a comment |
add a comment |
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$begingroup$
Given the information, $f$ can be seen as a probability density function of a r.v $Xin [0,1]$ such that $Bbb E[X]=a, Bbb E[X^2]=a^2$. Then, $textvar(X)=Bbb E[X^2]-Bbb E[X]^2=0$, which implies $Bbb P(X=a)=1$, but then $X$ cannot have a continuous pdf.
$endgroup$
– Song
Mar 22 at 4:09