Mixing
Does Cauchy-Schwarz imply $|x^Ty| leq |x|_p|y|_p$ for any $p geq 1$?
$begingroup$
Given $x,y in mathbb{R}^n$, the Cauchy Schwarz inequality states,
$|x^Ty| leq |x|_2|y|_2$
And for non-Euclidean (norms other than $l_2$), we have,
$|x^Ty| leq |x|_p|y|_q$ where $|cdot|_q$ is the dual norm of $|cdot|_p$ (I remember $p,q$ is related by $1/p+1/q = 1$ or something like that)
For example,
$|x^Ty| leq |x|_1|y|_infty$
My question is:
Does Cauchy-Schwarz imply $|x^Ty| leq |x|_p|y|_p$ for any $p
geq 1$?
I feel like this should hold, based on the fact that all the norms are equivalent. In the previous example, we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ (I can never remind the exact number)
Then it must be true that,
$|x^Ty| leq |x|_1|y|_1$
Is there a flaw in my reasoning?
inequality vector-spaces convex-analysis norm cauchy-schwarz-inequality
edited Jan 20 at 8:17
max_zorn
3,40061329
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
$endgroup$
add a comment |
$begingroup$
Given $x,y in mathbb{R}^n$, the Cauchy Schwarz inequality states,
$|x^Ty| leq |x|_2|y|_2$
And for non-Euclidean (norms other than $l_2$), we have,
$|x^Ty| leq |x|_p|y|_q$ where $|cdot|_q$ is the dual norm of $|cdot|_p$ (I remember $p,q$ is related by $1/p+1/q = 1$ or something like that)
For example,
$|x^Ty| leq |x|_1|y|_infty$
My question is:
Does Cauchy-Schwarz imply $|x^Ty| leq |x|_p|y|_p$ for any $p
geq 1$?
I feel like this should hold, based on the fact that all the norms are equivalent. In the previous example, we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ (I can never remind the exact number)
Then it must be true that,
$|x^Ty| leq |x|_1|y|_1$
Is there a flaw in my reasoning?
inequality vector-spaces convex-analysis norm cauchy-schwarz-inequality
edited Jan 20 at 8:17
max_zorn
3,40061329
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
$endgroup$
$begingroup$
We do have $|x^Ty| leq |x|_p |y|_p$ for $1 leq p leq 2$, but not for $p>2$.
$endgroup$
– Omnomnomnom
Jan 19 at 19:40
$begingroup$
The norms are equivalent indeed. But the constants in the norm comparison DO matter here.
$endgroup$
– J.F
Jan 19 at 19:47
add a comment |
$begingroup$
Given $x,y in mathbb{R}^n$, the Cauchy Schwarz inequality states,
$|x^Ty| leq |x|_2|y|_2$
And for non-Euclidean (norms other than $l_2$), we have,
$|x^Ty| leq |x|_p|y|_q$ where $|cdot|_q$ is the dual norm of $|cdot|_p$ (I remember $p,q$ is related by $1/p+1/q = 1$ or something like that)
For example,
$|x^Ty| leq |x|_1|y|_infty$
My question is:
Does Cauchy-Schwarz imply $|x^Ty| leq |x|_p|y|_p$ for any $p
geq 1$?
I feel like this should hold, based on the fact that all the norms are equivalent. In the previous example, we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ (I can never remind the exact number)
Then it must be true that,
$|x^Ty| leq |x|_1|y|_1$
Is there a flaw in my reasoning?
inequality vector-spaces convex-analysis norm cauchy-schwarz-inequality
edited Jan 20 at 8:17
max_zorn
3,40061329
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
$endgroup$
Given $x,y in mathbb{R}^n$, the Cauchy Schwarz inequality states,
$|x^Ty| leq |x|_2|y|_2$
And for non-Euclidean (norms other than $l_2$), we have,
$|x^Ty| leq |x|_p|y|_q$ where $|cdot|_q$ is the dual norm of $|cdot|_p$ (I remember $p,q$ is related by $1/p+1/q = 1$ or something like that)
For example,
$|x^Ty| leq |x|_1|y|_infty$
My question is:
Does Cauchy-Schwarz imply $|x^Ty| leq |x|_p|y|_p$ for any $p
geq 1$?
I feel like this should hold, based on the fact that all the norms are equivalent. In the previous example, we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ (I can never remind the exact number)
Then it must be true that,
$|x^Ty| leq |x|_1|y|_1$
Is there a flaw in my reasoning?
inequality vector-spaces convex-analysis norm cauchy-schwarz-inequality
inequality vector-spaces convex-analysis norm cauchy-schwarz-inequality
edited Jan 20 at 8:17
max_zorn
3,40061329
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
edited Jan 20 at 8:17
max_zorn
3,40061329
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
edited Jan 20 at 8:17
max_zorn
3,40061329
edited Jan 20 at 8:17
max_zorn
3,40061329
edited Jan 20 at 8:17
max_zorn
3,40061329
3,40061329
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
asked Jan 19 at 19:34
AåkonAåkon
4,86131758
4,86131758
$begingroup$
We do have $|x^Ty| leq |x|_p |y|_p$ for $1 leq p leq 2$, but not for $p>2$.
$endgroup$
– Omnomnomnom
Jan 19 at 19:40
$begingroup$
The norms are equivalent indeed. But the constants in the norm comparison DO matter here.
$endgroup$
– J.F
Jan 19 at 19:47
add a comment |
$begingroup$
We do have $|x^Ty| leq |x|_p |y|_p$ for $1 leq p leq 2$, but not for $p>2$.
$endgroup$
– Omnomnomnom
Jan 19 at 19:40
$begingroup$
The norms are equivalent indeed. But the constants in the norm comparison DO matter here.
$endgroup$
– J.F
Jan 19 at 19:47
$begingroup$
We do have $|x^Ty| leq |x|_p |y|_p$ for $1 leq p leq 2$, but not for $p>2$.
$endgroup$
– Omnomnomnom
Jan 19 at 19:40
$begingroup$
We do have $|x^Ty| leq |x|_p |y|_p$ for $1 leq p leq 2$, but not for $p>2$.
$endgroup$
– Omnomnomnom
Jan 19 at 19:40
$begingroup$
The norms are equivalent indeed. But the constants in the norm comparison DO matter here.
$endgroup$
– J.F
Jan 19 at 19:47
$begingroup$
The norms are equivalent indeed. But the constants in the norm comparison DO matter here.
$endgroup$
– J.F
Jan 19 at 19:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
... we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ ...
Well, actually we have $|y|_infty leq |y|_1$, and that implies
$$
|x^Ty| leq |x|_1|y|_infty le |x|_1|y|_1 , ,
$$
so that your reasoning is essentially correct for $p=1$.
For $p > 1$ the same approach works as long as the conjugate $q$ (defined by $frac 1p + frac 1q = 1$ ) satisfies $p le q$, that is, if $p le 2$. Then Hölder's inequality and the relation between $p$-norms gives
$$
|x^Ty| leq |x|_p|y|_q le |x|_p|y|_p , .
$$
The same reasoning does not work if $p > 2$, and in fact the desired estimate does not hold in that case. This can be seen by choosing $x = y ne 0$:
$$
|x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , .
$$
Summary:
$$
|x^Ty| leq Vert xVert_pVert yVert _p text{ for all } x, y in Bbb R^n
$$
holds if and only if $p le 2$.
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
$endgroup$
$begingroup$
Hi, I don't see how you obtained the inequality in the line $ |x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , . $ Why shouldn't the inequality be flipped?
$endgroup$
– Aåkon
Jan 19 at 22:37
$begingroup$
@Aåkon: The p-norm is decreasingin p: $p > 2$ implies $Vert x Vert_p < Vert x Vert_2$
$endgroup$
– Martin R
Jan 20 at 14:02
add a comment |
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1 Answer
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$begingroup$
... we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ ...
Well, actually we have $|y|_infty leq |y|_1$, and that implies
$$
|x^Ty| leq |x|_1|y|_infty le |x|_1|y|_1 , ,
$$
so that your reasoning is essentially correct for $p=1$.
For $p > 1$ the same approach works as long as the conjugate $q$ (defined by $frac 1p + frac 1q = 1$ ) satisfies $p le q$, that is, if $p le 2$. Then Hölder's inequality and the relation between $p$-norms gives
$$
|x^Ty| leq |x|_p|y|_q le |x|_p|y|_p , .
$$
The same reasoning does not work if $p > 2$, and in fact the desired estimate does not hold in that case. This can be seen by choosing $x = y ne 0$:
$$
|x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , .
$$
Summary:
$$
|x^Ty| leq Vert xVert_pVert yVert _p text{ for all } x, y in Bbb R^n
$$
holds if and only if $p le 2$.
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
$endgroup$
$begingroup$
Hi, I don't see how you obtained the inequality in the line $ |x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , . $ Why shouldn't the inequality be flipped?
$endgroup$
– Aåkon
Jan 19 at 22:37
$begingroup$
@Aåkon: The p-norm is decreasingin p: $p > 2$ implies $Vert x Vert_p < Vert x Vert_2$
$endgroup$
– Martin R
Jan 20 at 14:02
add a comment |
$begingroup$
... we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ ...
Well, actually we have $|y|_infty leq |y|_1$, and that implies
$$
|x^Ty| leq |x|_1|y|_infty le |x|_1|y|_1 , ,
$$
so that your reasoning is essentially correct for $p=1$.
For $p > 1$ the same approach works as long as the conjugate $q$ (defined by $frac 1p + frac 1q = 1$ ) satisfies $p le q$, that is, if $p le 2$. Then Hölder's inequality and the relation between $p$-norms gives
$$
|x^Ty| leq |x|_p|y|_q le |x|_p|y|_p , .
$$
The same reasoning does not work if $p > 2$, and in fact the desired estimate does not hold in that case. This can be seen by choosing $x = y ne 0$:
$$
|x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , .
$$
Summary:
$$
|x^Ty| leq Vert xVert_pVert yVert _p text{ for all } x, y in Bbb R^n
$$
holds if and only if $p le 2$.
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
$endgroup$
$begingroup$
Hi, I don't see how you obtained the inequality in the line $ |x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , . $ Why shouldn't the inequality be flipped?
$endgroup$
– Aåkon
Jan 19 at 22:37
$begingroup$
@Aåkon: The p-norm is decreasingin p: $p > 2$ implies $Vert x Vert_p < Vert x Vert_2$
$endgroup$
– Martin R
Jan 20 at 14:02
add a comment |
$begingroup$
... we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ ...
Well, actually we have $|y|_infty leq |y|_1$, and that implies
$$
|x^Ty| leq |x|_1|y|_infty le |x|_1|y|_1 , ,
$$
so that your reasoning is essentially correct for $p=1$.
For $p > 1$ the same approach works as long as the conjugate $q$ (defined by $frac 1p + frac 1q = 1$ ) satisfies $p le q$, that is, if $p le 2$. Then Hölder's inequality and the relation between $p$-norms gives
$$
|x^Ty| leq |x|_p|y|_q le |x|_p|y|_p , .
$$
The same reasoning does not work if $p > 2$, and in fact the desired estimate does not hold in that case. This can be seen by choosing $x = y ne 0$:
$$
|x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , .
$$
Summary:
$$
|x^Ty| leq Vert xVert_pVert yVert _p text{ for all } x, y in Bbb R^n
$$
holds if and only if $p le 2$.
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
$endgroup$
... we know that $|y|_infty leq alpha |y|_1$ with some constant $alpha$ ...
Well, actually we have $|y|_infty leq |y|_1$, and that implies
$$
|x^Ty| leq |x|_1|y|_infty le |x|_1|y|_1 , ,
$$
so that your reasoning is essentially correct for $p=1$.
For $p > 1$ the same approach works as long as the conjugate $q$ (defined by $frac 1p + frac 1q = 1$ ) satisfies $p le q$, that is, if $p le 2$. Then Hölder's inequality and the relation between $p$-norms gives
$$
|x^Ty| leq |x|_p|y|_q le |x|_p|y|_p , .
$$
The same reasoning does not work if $p > 2$, and in fact the desired estimate does not hold in that case. This can be seen by choosing $x = y ne 0$:
$$
|x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , .
$$
Summary:
$$
|x^Ty| leq Vert xVert_pVert yVert _p text{ for all } x, y in Bbb R^n
$$
holds if and only if $p le 2$.
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
edited Jan 19 at 21:35
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
answered Jan 19 at 21:22
Martin RMartin R
29.6k33558
29.6k33558
$begingroup$
Hi, I don't see how you obtained the inequality in the line $ |x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , . $ Why shouldn't the inequality be flipped?
$endgroup$
– Aåkon
Jan 19 at 22:37
$begingroup$
@Aåkon: The p-norm is decreasingin p: $p > 2$ implies $Vert x Vert_p < Vert x Vert_2$
$endgroup$
– Martin R
Jan 20 at 14:02
add a comment |
$begingroup$
Hi, I don't see how you obtained the inequality in the line $ |x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , . $ Why shouldn't the inequality be flipped?
$endgroup$
– Aåkon
Jan 19 at 22:37
$begingroup$
@Aåkon: The p-norm is decreasingin p: $p > 2$ implies $Vert x Vert_p < Vert x Vert_2$
$endgroup$
– Martin R
Jan 20 at 14:02
$begingroup$
Hi, I don't see how you obtained the inequality in the line $ |x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , . $ Why shouldn't the inequality be flipped?
$endgroup$
– Aåkon
Jan 19 at 22:37
$begingroup$
Hi, I don't see how you obtained the inequality in the line $ |x^Ty| = Vert x Vert_2^2 > Vert x Vert_p^2 = |x|_p|y|_p , . $ Why shouldn't the inequality be flipped?
$endgroup$
– Aåkon
Jan 19 at 22:37
$begingroup$
@Aåkon: The p-norm is decreasingin p: $p > 2$ implies $Vert x Vert_p < Vert x Vert_2$
$endgroup$
– Martin R
Jan 20 at 14:02
$begingroup$
@Aåkon: The p-norm is decreasingin p: $p > 2$ implies $Vert x Vert_p < Vert x Vert_2$
$endgroup$
– Martin R
Jan 20 at 14:02
add a comment |
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$begingroup$
We do have $|x^Ty| leq |x|_p |y|_p$ for $1 leq p leq 2$, but not for $p>2$.
$endgroup$
– Omnomnomnom
Jan 19 at 19:40
$begingroup$
The norms are equivalent indeed. But the constants in the norm comparison DO matter here.
$endgroup$
– J.F
Jan 19 at 19:47