Mixing
Sums of vectors in $l_p$
$begingroup$
Suppose $x$ and $y$ are two vectors in $l_p$ (where $1leq p<infty$) such that $||x||=||y||=1$. Can we find a complex scalar $alpha$, with $|alpha|<1$ such that $||x+alpha y||>1$. I don't know the answer to this even in finite dimensional $l_p^n$. For example, in $l_1^n$ thsi can be rephrased as:
Given $displaystyle sum_{i=1}^n|a_i|=sum_{i=1}^n|b_i|=1$, can we find $|alpha|<1$ such that $displaystyle sum_{i=1}^n|a_i+alpha b_i|>1$?
real-analysis complex-analysis functional-analysis banach-spaces
asked Jan 30 at 1:12
MarkusMarkus
933511
$endgroup$
add a comment |
$begingroup$
Suppose $x$ and $y$ are two vectors in $l_p$ (where $1leq p<infty$) such that $||x||=||y||=1$. Can we find a complex scalar $alpha$, with $|alpha|<1$ such that $||x+alpha y||>1$. I don't know the answer to this even in finite dimensional $l_p^n$. For example, in $l_1^n$ thsi can be rephrased as:
Given $displaystyle sum_{i=1}^n|a_i|=sum_{i=1}^n|b_i|=1$, can we find $|alpha|<1$ such that $displaystyle sum_{i=1}^n|a_i+alpha b_i|>1$?
real-analysis complex-analysis functional-analysis banach-spaces
asked Jan 30 at 1:12
MarkusMarkus
933511
$endgroup$
add a comment |
$begingroup$
Suppose $x$ and $y$ are two vectors in $l_p$ (where $1leq p<infty$) such that $||x||=||y||=1$. Can we find a complex scalar $alpha$, with $|alpha|<1$ such that $||x+alpha y||>1$. I don't know the answer to this even in finite dimensional $l_p^n$. For example, in $l_1^n$ thsi can be rephrased as:
Given $displaystyle sum_{i=1}^n|a_i|=sum_{i=1}^n|b_i|=1$, can we find $|alpha|<1$ such that $displaystyle sum_{i=1}^n|a_i+alpha b_i|>1$?
real-analysis complex-analysis functional-analysis banach-spaces
asked Jan 30 at 1:12
MarkusMarkus
933511
$endgroup$
Suppose $x$ and $y$ are two vectors in $l_p$ (where $1leq p<infty$) such that $||x||=||y||=1$. Can we find a complex scalar $alpha$, with $|alpha|<1$ such that $||x+alpha y||>1$. I don't know the answer to this even in finite dimensional $l_p^n$. For example, in $l_1^n$ thsi can be rephrased as:
Given $displaystyle sum_{i=1}^n|a_i|=sum_{i=1}^n|b_i|=1$, can we find $|alpha|<1$ such that $displaystyle sum_{i=1}^n|a_i+alpha b_i|>1$?
real-analysis complex-analysis functional-analysis banach-spaces
real-analysis complex-analysis functional-analysis banach-spaces
asked Jan 30 at 1:12
MarkusMarkus
933511
asked Jan 30 at 1:12
MarkusMarkus
933511
asked Jan 30 at 1:12
MarkusMarkus
933511
asked Jan 30 at 1:12
MarkusMarkus
933511
asked Jan 30 at 1:12
MarkusMarkus
933511
933511
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose $X$ is an arbitrary normed linear space, and $|x| = 1$. Then, at least one of $|x + y|$ and $|x - y|$ must be greater than or equal to $1$, as
$$|x + y| + |x - y| ge |2x| = 2.$$
To get strict inequality, assume further that $y neq 0$ and $X$ is a strictly convex space, meaning that the unit sphere of $X$ contains no (non-degenerate) line segments (e.g. $l_p$ spaces for $1 < p < infty$). Equivalently, if $v, w$ have norm $1$, then $frac{v + w}{2}$ has norm strictly less than $1$.
If $|x + y|, |x - y| le 1$, then
$$2 ge |x + y| + |x - y| ge |2x| = 2,$$
which means that $|x + y| = |x - y| = 1$. Note that
$$x = frac{(x + y) + (x - y)}{2}.$$
This contradicts strict convexity. Thus, in a strictly convex space, $|x + y| > 1$ or $|x - y| > 1$.
EDIT: This property fails if and only if $X$ has a subspace that is isometrically isomorphic to $ell^2_infty$. Fix $|x| = |y| = 1$, and suppose $|x + alpha y| le 1$ for all $|alpha| < 1$.
From the above methods, we can conclude that $|x + alpha y| = 1$ for all $|alpha| < 1$, as if $|x + alpha y| < 1$, then $|x - alpha y| > 1$. By continuity, this implies $|x + alpha y| = 1$ for $|alpha| = 1$ too.
Now, if $|alpha| = 1$, then $|alpha^{-1}| = 1$, so $|y + alpha x| = |alpha| |x + alpha^{-1}y| = 1$. Note that $y$ is the midpoint of $y + alpha x$ and $y - alpha x$, all of which have norm $1$, so every point on the line segment between $y + alpha x$ and $y - alpha x$ have norm $1$. That is, if $|alpha| le 1$, we have $|y + alpha x| = |x + alpha y| = 1$.
Now, it's easy to see that $|alpha x + beta y| = max{|alpha|, |beta|}$, by simply dividing by $alpha$ or $beta$; whichever has largest modulus. Our linear isometry is defined by $x mapsto (1, 0)$ and $y mapsto (0, 1)$ (note that $x neq y$, as $x = y$ would imply that $|x + 2y| = 3$, not $2$).
So, all we need to do is show that $ell_1$ does not contain such a subspace (this will also prove the property for $ell_1^n$, as it can be identified with a subspace of $ell_1$). Certainly in the real case, such a subspace does exist: $ell_1^2$ (map $(1/2, 1/2)$ to $(1, 0)$ and $(1/2, -1/2)$ to $(0, 1)$), so the property fails. For the complex case, I don't know.
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
$endgroup$
1
$begingroup$
I believe the question requires $|alpha| < 1$.
$endgroup$
– Timothy Hedgeworth
Jan 30 at 1:40
1
$begingroup$
@TimothyHedgeworth Note that there's no norm restriction on $y$, so it doesn't matter.
$endgroup$
– Theo Bendit
Jan 30 at 1:43
1
$begingroup$
But $l_1$ is not strictly convex. Can you get this in $l_1$ as well, this is in fact the case I am mostly interested in.?
$endgroup$
– Theo
Jan 30 at 2:26
$begingroup$
@Theo I'll have to think about it.
$endgroup$
– Theo Bendit
Jan 30 at 2:30
$begingroup$
@Theo I can show that this property fails if and only if $X$ contains a subspace isometrically isomorphic to $ell_infty^2$. I'm not certain how to show that no such subspace exists in $ell_1$ (over the complex numbers).
$endgroup$
– Theo Bendit
Jan 30 at 4:17
|
show 3 more comments
$begingroup$
If $1<p<infty$ we've seen it's just strict convexity. The result is false for $p=infty$. For $p=1$ there's a sort of complex version of strict convexity; although the unit sphere of $L^1$ does contain line segments it does not contain any (flat) complex disks:
Lemma. If $z,winBbb C$ and $wne0$ then $frac1{2pi}int_0^{2pi}|z+e^{it}w|,dt>|z|$.
Proof: This is clear for $z=0$. If $zne0$ then wlog $z=1$ and then wlog $w=r>0$. The integral is equal to $frac1{2pi}int_0^{2pi}phi(t),dt$, where $$phi(t)=frac 12(|1+e^{it}r|+|1-e^{it}r|).$$The triangle inequality shows that $phi(t)ge1$, and it's clear that $phi(pi/2)>1$, hence the integral is greater than $1$ since $phi$ is continuous.
Cor. If $f,gin L^1(mu)$ and $gne0$ then $frac1{2pi}int_0^{2pi}||f+e^{it}g||_1,dt>||f||_1$.
(So there exists $alpha$ with $|alpha|=1$ and $||f+alpha g||_1>||f||_1$, and hence you can also get $|alpha|<1$.)
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Suppose $X$ is an arbitrary normed linear space, and $|x| = 1$. Then, at least one of $|x + y|$ and $|x - y|$ must be greater than or equal to $1$, as
$$|x + y| + |x - y| ge |2x| = 2.$$
To get strict inequality, assume further that $y neq 0$ and $X$ is a strictly convex space, meaning that the unit sphere of $X$ contains no (non-degenerate) line segments (e.g. $l_p$ spaces for $1 < p < infty$). Equivalently, if $v, w$ have norm $1$, then $frac{v + w}{2}$ has norm strictly less than $1$.
If $|x + y|, |x - y| le 1$, then
$$2 ge |x + y| + |x - y| ge |2x| = 2,$$
which means that $|x + y| = |x - y| = 1$. Note that
$$x = frac{(x + y) + (x - y)}{2}.$$
This contradicts strict convexity. Thus, in a strictly convex space, $|x + y| > 1$ or $|x - y| > 1$.
EDIT: This property fails if and only if $X$ has a subspace that is isometrically isomorphic to $ell^2_infty$. Fix $|x| = |y| = 1$, and suppose $|x + alpha y| le 1$ for all $|alpha| < 1$.
From the above methods, we can conclude that $|x + alpha y| = 1$ for all $|alpha| < 1$, as if $|x + alpha y| < 1$, then $|x - alpha y| > 1$. By continuity, this implies $|x + alpha y| = 1$ for $|alpha| = 1$ too.
Now, if $|alpha| = 1$, then $|alpha^{-1}| = 1$, so $|y + alpha x| = |alpha| |x + alpha^{-1}y| = 1$. Note that $y$ is the midpoint of $y + alpha x$ and $y - alpha x$, all of which have norm $1$, so every point on the line segment between $y + alpha x$ and $y - alpha x$ have norm $1$. That is, if $|alpha| le 1$, we have $|y + alpha x| = |x + alpha y| = 1$.
Now, it's easy to see that $|alpha x + beta y| = max{|alpha|, |beta|}$, by simply dividing by $alpha$ or $beta$; whichever has largest modulus. Our linear isometry is defined by $x mapsto (1, 0)$ and $y mapsto (0, 1)$ (note that $x neq y$, as $x = y$ would imply that $|x + 2y| = 3$, not $2$).
So, all we need to do is show that $ell_1$ does not contain such a subspace (this will also prove the property for $ell_1^n$, as it can be identified with a subspace of $ell_1$). Certainly in the real case, such a subspace does exist: $ell_1^2$ (map $(1/2, 1/2)$ to $(1, 0)$ and $(1/2, -1/2)$ to $(0, 1)$), so the property fails. For the complex case, I don't know.
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
$endgroup$
1
$begingroup$
I believe the question requires $|alpha| < 1$.
$endgroup$
– Timothy Hedgeworth
Jan 30 at 1:40
1
$begingroup$
@TimothyHedgeworth Note that there's no norm restriction on $y$, so it doesn't matter.
$endgroup$
– Theo Bendit
Jan 30 at 1:43
1
$begingroup$
But $l_1$ is not strictly convex. Can you get this in $l_1$ as well, this is in fact the case I am mostly interested in.?
$endgroup$
– Theo
Jan 30 at 2:26
$begingroup$
@Theo I'll have to think about it.
$endgroup$
– Theo Bendit
Jan 30 at 2:30
$begingroup$
@Theo I can show that this property fails if and only if $X$ contains a subspace isometrically isomorphic to $ell_infty^2$. I'm not certain how to show that no such subspace exists in $ell_1$ (over the complex numbers).
$endgroup$
– Theo Bendit
Jan 30 at 4:17
|
show 3 more comments
$begingroup$
Suppose $X$ is an arbitrary normed linear space, and $|x| = 1$. Then, at least one of $|x + y|$ and $|x - y|$ must be greater than or equal to $1$, as
$$|x + y| + |x - y| ge |2x| = 2.$$
To get strict inequality, assume further that $y neq 0$ and $X$ is a strictly convex space, meaning that the unit sphere of $X$ contains no (non-degenerate) line segments (e.g. $l_p$ spaces for $1 < p < infty$). Equivalently, if $v, w$ have norm $1$, then $frac{v + w}{2}$ has norm strictly less than $1$.
If $|x + y|, |x - y| le 1$, then
$$2 ge |x + y| + |x - y| ge |2x| = 2,$$
which means that $|x + y| = |x - y| = 1$. Note that
$$x = frac{(x + y) + (x - y)}{2}.$$
This contradicts strict convexity. Thus, in a strictly convex space, $|x + y| > 1$ or $|x - y| > 1$.
EDIT: This property fails if and only if $X$ has a subspace that is isometrically isomorphic to $ell^2_infty$. Fix $|x| = |y| = 1$, and suppose $|x + alpha y| le 1$ for all $|alpha| < 1$.
From the above methods, we can conclude that $|x + alpha y| = 1$ for all $|alpha| < 1$, as if $|x + alpha y| < 1$, then $|x - alpha y| > 1$. By continuity, this implies $|x + alpha y| = 1$ for $|alpha| = 1$ too.
Now, if $|alpha| = 1$, then $|alpha^{-1}| = 1$, so $|y + alpha x| = |alpha| |x + alpha^{-1}y| = 1$. Note that $y$ is the midpoint of $y + alpha x$ and $y - alpha x$, all of which have norm $1$, so every point on the line segment between $y + alpha x$ and $y - alpha x$ have norm $1$. That is, if $|alpha| le 1$, we have $|y + alpha x| = |x + alpha y| = 1$.
Now, it's easy to see that $|alpha x + beta y| = max{|alpha|, |beta|}$, by simply dividing by $alpha$ or $beta$; whichever has largest modulus. Our linear isometry is defined by $x mapsto (1, 0)$ and $y mapsto (0, 1)$ (note that $x neq y$, as $x = y$ would imply that $|x + 2y| = 3$, not $2$).
So, all we need to do is show that $ell_1$ does not contain such a subspace (this will also prove the property for $ell_1^n$, as it can be identified with a subspace of $ell_1$). Certainly in the real case, such a subspace does exist: $ell_1^2$ (map $(1/2, 1/2)$ to $(1, 0)$ and $(1/2, -1/2)$ to $(0, 1)$), so the property fails. For the complex case, I don't know.
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
$endgroup$
1
$begingroup$
I believe the question requires $|alpha| < 1$.
$endgroup$
– Timothy Hedgeworth
Jan 30 at 1:40
1
$begingroup$
@TimothyHedgeworth Note that there's no norm restriction on $y$, so it doesn't matter.
$endgroup$
– Theo Bendit
Jan 30 at 1:43
1
$begingroup$
But $l_1$ is not strictly convex. Can you get this in $l_1$ as well, this is in fact the case I am mostly interested in.?
$endgroup$
– Theo
Jan 30 at 2:26
$begingroup$
@Theo I'll have to think about it.
$endgroup$
– Theo Bendit
Jan 30 at 2:30
$begingroup$
@Theo I can show that this property fails if and only if $X$ contains a subspace isometrically isomorphic to $ell_infty^2$. I'm not certain how to show that no such subspace exists in $ell_1$ (over the complex numbers).
$endgroup$
– Theo Bendit
Jan 30 at 4:17
|
show 3 more comments
$begingroup$
Suppose $X$ is an arbitrary normed linear space, and $|x| = 1$. Then, at least one of $|x + y|$ and $|x - y|$ must be greater than or equal to $1$, as
$$|x + y| + |x - y| ge |2x| = 2.$$
To get strict inequality, assume further that $y neq 0$ and $X$ is a strictly convex space, meaning that the unit sphere of $X$ contains no (non-degenerate) line segments (e.g. $l_p$ spaces for $1 < p < infty$). Equivalently, if $v, w$ have norm $1$, then $frac{v + w}{2}$ has norm strictly less than $1$.
If $|x + y|, |x - y| le 1$, then
$$2 ge |x + y| + |x - y| ge |2x| = 2,$$
which means that $|x + y| = |x - y| = 1$. Note that
$$x = frac{(x + y) + (x - y)}{2}.$$
This contradicts strict convexity. Thus, in a strictly convex space, $|x + y| > 1$ or $|x - y| > 1$.
EDIT: This property fails if and only if $X$ has a subspace that is isometrically isomorphic to $ell^2_infty$. Fix $|x| = |y| = 1$, and suppose $|x + alpha y| le 1$ for all $|alpha| < 1$.
From the above methods, we can conclude that $|x + alpha y| = 1$ for all $|alpha| < 1$, as if $|x + alpha y| < 1$, then $|x - alpha y| > 1$. By continuity, this implies $|x + alpha y| = 1$ for $|alpha| = 1$ too.
Now, if $|alpha| = 1$, then $|alpha^{-1}| = 1$, so $|y + alpha x| = |alpha| |x + alpha^{-1}y| = 1$. Note that $y$ is the midpoint of $y + alpha x$ and $y - alpha x$, all of which have norm $1$, so every point on the line segment between $y + alpha x$ and $y - alpha x$ have norm $1$. That is, if $|alpha| le 1$, we have $|y + alpha x| = |x + alpha y| = 1$.
Now, it's easy to see that $|alpha x + beta y| = max{|alpha|, |beta|}$, by simply dividing by $alpha$ or $beta$; whichever has largest modulus. Our linear isometry is defined by $x mapsto (1, 0)$ and $y mapsto (0, 1)$ (note that $x neq y$, as $x = y$ would imply that $|x + 2y| = 3$, not $2$).
So, all we need to do is show that $ell_1$ does not contain such a subspace (this will also prove the property for $ell_1^n$, as it can be identified with a subspace of $ell_1$). Certainly in the real case, such a subspace does exist: $ell_1^2$ (map $(1/2, 1/2)$ to $(1, 0)$ and $(1/2, -1/2)$ to $(0, 1)$), so the property fails. For the complex case, I don't know.
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
$endgroup$
Suppose $X$ is an arbitrary normed linear space, and $|x| = 1$. Then, at least one of $|x + y|$ and $|x - y|$ must be greater than or equal to $1$, as
$$|x + y| + |x - y| ge |2x| = 2.$$
To get strict inequality, assume further that $y neq 0$ and $X$ is a strictly convex space, meaning that the unit sphere of $X$ contains no (non-degenerate) line segments (e.g. $l_p$ spaces for $1 < p < infty$). Equivalently, if $v, w$ have norm $1$, then $frac{v + w}{2}$ has norm strictly less than $1$.
If $|x + y|, |x - y| le 1$, then
$$2 ge |x + y| + |x - y| ge |2x| = 2,$$
which means that $|x + y| = |x - y| = 1$. Note that
$$x = frac{(x + y) + (x - y)}{2}.$$
This contradicts strict convexity. Thus, in a strictly convex space, $|x + y| > 1$ or $|x - y| > 1$.
EDIT: This property fails if and only if $X$ has a subspace that is isometrically isomorphic to $ell^2_infty$. Fix $|x| = |y| = 1$, and suppose $|x + alpha y| le 1$ for all $|alpha| < 1$.
From the above methods, we can conclude that $|x + alpha y| = 1$ for all $|alpha| < 1$, as if $|x + alpha y| < 1$, then $|x - alpha y| > 1$. By continuity, this implies $|x + alpha y| = 1$ for $|alpha| = 1$ too.
Now, if $|alpha| = 1$, then $|alpha^{-1}| = 1$, so $|y + alpha x| = |alpha| |x + alpha^{-1}y| = 1$. Note that $y$ is the midpoint of $y + alpha x$ and $y - alpha x$, all of which have norm $1$, so every point on the line segment between $y + alpha x$ and $y - alpha x$ have norm $1$. That is, if $|alpha| le 1$, we have $|y + alpha x| = |x + alpha y| = 1$.
Now, it's easy to see that $|alpha x + beta y| = max{|alpha|, |beta|}$, by simply dividing by $alpha$ or $beta$; whichever has largest modulus. Our linear isometry is defined by $x mapsto (1, 0)$ and $y mapsto (0, 1)$ (note that $x neq y$, as $x = y$ would imply that $|x + 2y| = 3$, not $2$).
So, all we need to do is show that $ell_1$ does not contain such a subspace (this will also prove the property for $ell_1^n$, as it can be identified with a subspace of $ell_1$). Certainly in the real case, such a subspace does exist: $ell_1^2$ (map $(1/2, 1/2)$ to $(1, 0)$ and $(1/2, -1/2)$ to $(0, 1)$), so the property fails. For the complex case, I don't know.
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
edited Jan 30 at 4:24
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
answered Jan 30 at 1:31
Theo BenditTheo Bendit
20.1k12354
20.1k12354
1
$begingroup$
I believe the question requires $|alpha| < 1$.
$endgroup$
– Timothy Hedgeworth
Jan 30 at 1:40
1
$begingroup$
@TimothyHedgeworth Note that there's no norm restriction on $y$, so it doesn't matter.
$endgroup$
– Theo Bendit
Jan 30 at 1:43
1
$begingroup$
But $l_1$ is not strictly convex. Can you get this in $l_1$ as well, this is in fact the case I am mostly interested in.?
$endgroup$
– Theo
Jan 30 at 2:26
$begingroup$
@Theo I'll have to think about it.
$endgroup$
– Theo Bendit
Jan 30 at 2:30
$begingroup$
@Theo I can show that this property fails if and only if $X$ contains a subspace isometrically isomorphic to $ell_infty^2$. I'm not certain how to show that no such subspace exists in $ell_1$ (over the complex numbers).
$endgroup$
– Theo Bendit
Jan 30 at 4:17
|
show 3 more comments
1
$begingroup$
I believe the question requires $|alpha| < 1$.
$endgroup$
– Timothy Hedgeworth
Jan 30 at 1:40
1
$begingroup$
@TimothyHedgeworth Note that there's no norm restriction on $y$, so it doesn't matter.
$endgroup$
– Theo Bendit
Jan 30 at 1:43
1
$begingroup$
But $l_1$ is not strictly convex. Can you get this in $l_1$ as well, this is in fact the case I am mostly interested in.?
$endgroup$
– Theo
Jan 30 at 2:26
$begingroup$
@Theo I'll have to think about it.
$endgroup$
– Theo Bendit
Jan 30 at 2:30
$begingroup$
@Theo I can show that this property fails if and only if $X$ contains a subspace isometrically isomorphic to $ell_infty^2$. I'm not certain how to show that no such subspace exists in $ell_1$ (over the complex numbers).
$endgroup$
– Theo Bendit
Jan 30 at 4:17
1
1
$begingroup$
I believe the question requires $|alpha| < 1$.
$endgroup$
– Timothy Hedgeworth
Jan 30 at 1:40
$begingroup$
I believe the question requires $|alpha| < 1$.
$endgroup$
– Timothy Hedgeworth
Jan 30 at 1:40
1
1
$begingroup$
@TimothyHedgeworth Note that there's no norm restriction on $y$, so it doesn't matter.
$endgroup$
– Theo Bendit
Jan 30 at 1:43
$begingroup$
@TimothyHedgeworth Note that there's no norm restriction on $y$, so it doesn't matter.
$endgroup$
– Theo Bendit
Jan 30 at 1:43
1
1
$begingroup$
But $l_1$ is not strictly convex. Can you get this in $l_1$ as well, this is in fact the case I am mostly interested in.?
$endgroup$
– Theo
Jan 30 at 2:26
$begingroup$
But $l_1$ is not strictly convex. Can you get this in $l_1$ as well, this is in fact the case I am mostly interested in.?
$endgroup$
– Theo
Jan 30 at 2:26
$begingroup$
@Theo I'll have to think about it.
$endgroup$
– Theo Bendit
Jan 30 at 2:30
$begingroup$
@Theo I'll have to think about it.
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– Theo Bendit
Jan 30 at 2:30
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@Theo I can show that this property fails if and only if $X$ contains a subspace isometrically isomorphic to $ell_infty^2$. I'm not certain how to show that no such subspace exists in $ell_1$ (over the complex numbers).
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– Theo Bendit
Jan 30 at 4:17
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@Theo I can show that this property fails if and only if $X$ contains a subspace isometrically isomorphic to $ell_infty^2$. I'm not certain how to show that no such subspace exists in $ell_1$ (over the complex numbers).
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– Theo Bendit
Jan 30 at 4:17
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show 3 more comments
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If $1<p<infty$ we've seen it's just strict convexity. The result is false for $p=infty$. For $p=1$ there's a sort of complex version of strict convexity; although the unit sphere of $L^1$ does contain line segments it does not contain any (flat) complex disks:
Lemma. If $z,winBbb C$ and $wne0$ then $frac1{2pi}int_0^{2pi}|z+e^{it}w|,dt>|z|$.
Proof: This is clear for $z=0$. If $zne0$ then wlog $z=1$ and then wlog $w=r>0$. The integral is equal to $frac1{2pi}int_0^{2pi}phi(t),dt$, where $$phi(t)=frac 12(|1+e^{it}r|+|1-e^{it}r|).$$The triangle inequality shows that $phi(t)ge1$, and it's clear that $phi(pi/2)>1$, hence the integral is greater than $1$ since $phi$ is continuous.
Cor. If $f,gin L^1(mu)$ and $gne0$ then $frac1{2pi}int_0^{2pi}||f+e^{it}g||_1,dt>||f||_1$.
(So there exists $alpha$ with $|alpha|=1$ and $||f+alpha g||_1>||f||_1$, and hence you can also get $|alpha|<1$.)
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
$begingroup$
If $1<p<infty$ we've seen it's just strict convexity. The result is false for $p=infty$. For $p=1$ there's a sort of complex version of strict convexity; although the unit sphere of $L^1$ does contain line segments it does not contain any (flat) complex disks:
Lemma. If $z,winBbb C$ and $wne0$ then $frac1{2pi}int_0^{2pi}|z+e^{it}w|,dt>|z|$.
Proof: This is clear for $z=0$. If $zne0$ then wlog $z=1$ and then wlog $w=r>0$. The integral is equal to $frac1{2pi}int_0^{2pi}phi(t),dt$, where $$phi(t)=frac 12(|1+e^{it}r|+|1-e^{it}r|).$$The triangle inequality shows that $phi(t)ge1$, and it's clear that $phi(pi/2)>1$, hence the integral is greater than $1$ since $phi$ is continuous.
Cor. If $f,gin L^1(mu)$ and $gne0$ then $frac1{2pi}int_0^{2pi}||f+e^{it}g||_1,dt>||f||_1$.
(So there exists $alpha$ with $|alpha|=1$ and $||f+alpha g||_1>||f||_1$, and hence you can also get $|alpha|<1$.)
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
add a comment |
$begingroup$
If $1<p<infty$ we've seen it's just strict convexity. The result is false for $p=infty$. For $p=1$ there's a sort of complex version of strict convexity; although the unit sphere of $L^1$ does contain line segments it does not contain any (flat) complex disks:
Lemma. If $z,winBbb C$ and $wne0$ then $frac1{2pi}int_0^{2pi}|z+e^{it}w|,dt>|z|$.
Proof: This is clear for $z=0$. If $zne0$ then wlog $z=1$ and then wlog $w=r>0$. The integral is equal to $frac1{2pi}int_0^{2pi}phi(t),dt$, where $$phi(t)=frac 12(|1+e^{it}r|+|1-e^{it}r|).$$The triangle inequality shows that $phi(t)ge1$, and it's clear that $phi(pi/2)>1$, hence the integral is greater than $1$ since $phi$ is continuous.
Cor. If $f,gin L^1(mu)$ and $gne0$ then $frac1{2pi}int_0^{2pi}||f+e^{it}g||_1,dt>||f||_1$.
(So there exists $alpha$ with $|alpha|=1$ and $||f+alpha g||_1>||f||_1$, and hence you can also get $|alpha|<1$.)
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
If $1<p<infty$ we've seen it's just strict convexity. The result is false for $p=infty$. For $p=1$ there's a sort of complex version of strict convexity; although the unit sphere of $L^1$ does contain line segments it does not contain any (flat) complex disks:
Lemma. If $z,winBbb C$ and $wne0$ then $frac1{2pi}int_0^{2pi}|z+e^{it}w|,dt>|z|$.
Proof: This is clear for $z=0$. If $zne0$ then wlog $z=1$ and then wlog $w=r>0$. The integral is equal to $frac1{2pi}int_0^{2pi}phi(t),dt$, where $$phi(t)=frac 12(|1+e^{it}r|+|1-e^{it}r|).$$The triangle inequality shows that $phi(t)ge1$, and it's clear that $phi(pi/2)>1$, hence the integral is greater than $1$ since $phi$ is continuous.
Cor. If $f,gin L^1(mu)$ and $gne0$ then $frac1{2pi}int_0^{2pi}||f+e^{it}g||_1,dt>||f||_1$.
(So there exists $alpha$ with $|alpha|=1$ and $||f+alpha g||_1>||f||_1$, and hence you can also get $|alpha|<1$.)
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
edited Jan 30 at 15:22
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 30 at 15:06
David C. UllrichDavid C. Ullrich
61.6k43995
61.6k43995
add a comment |
add a comment |
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