Reason for the term “smooth”
$begingroup$
A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?
functional-analysis banach-spaces normed-spaces
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
$endgroup$
add a comment |
$begingroup$
A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?
functional-analysis banach-spaces normed-spaces
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
$endgroup$
add a comment |
$begingroup$
A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?
functional-analysis banach-spaces normed-spaces
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
$endgroup$
A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?
functional-analysis banach-spaces normed-spaces
functional-analysis banach-spaces normed-spaces
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
asked Nov 28 '18 at 8:31
InfinityInfinity
629514
629514
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.
Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
It can be shown, that there are points $xin X$ such that there is more than one functional $f$
with $|f|=1$ and $f(x)=|x|=1$.
(For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)
For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
And it is possible to show that for each $xin X$ there is exactly one functional $f$
with $|f|=f(x)=|x|=1$.
I hope this is sufficient motivation for the term "smooth" for a normed space.
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
$endgroup$
add a comment |
$begingroup$
This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
Indeed, the set
$$
{ fin X^* mid |f| le 1 text{ and } f(x) = |x|}$$
coincides with the convex subdifferential of $F$ at $x$.
If this subdifferential is a singleton, then $F$ is differentiable.
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
$endgroup$
$begingroup$
$F'(x)=f$ is a typo?
$endgroup$
– David C. Ullrich
Nov 28 '18 at 15:41
$begingroup$
No, the derivative of $F$ at $x$ is the linear function $f$.
$endgroup$
– gerw
Nov 28 '18 at 18:01
$begingroup$
The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
$endgroup$
– David C. Ullrich
Nov 28 '18 at 18:31
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.
Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
It can be shown, that there are points $xin X$ such that there is more than one functional $f$
with $|f|=1$ and $f(x)=|x|=1$.
(For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)
For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
And it is possible to show that for each $xin X$ there is exactly one functional $f$
with $|f|=f(x)=|x|=1$.
I hope this is sufficient motivation for the term "smooth" for a normed space.
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
$endgroup$
add a comment |
$begingroup$
Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.
Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
It can be shown, that there are points $xin X$ such that there is more than one functional $f$
with $|f|=1$ and $f(x)=|x|=1$.
(For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)
For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
And it is possible to show that for each $xin X$ there is exactly one functional $f$
with $|f|=f(x)=|x|=1$.
I hope this is sufficient motivation for the term "smooth" for a normed space.
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
$endgroup$
add a comment |
$begingroup$
Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.
Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
It can be shown, that there are points $xin X$ such that there is more than one functional $f$
with $|f|=1$ and $f(x)=|x|=1$.
(For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)
For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
And it is possible to show that for each $xin X$ there is exactly one functional $f$
with $|f|=f(x)=|x|=1$.
I hope this is sufficient motivation for the term "smooth" for a normed space.
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
$endgroup$
Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.
Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
It can be shown, that there are points $xin X$ such that there is more than one functional $f$
with $|f|=1$ and $f(x)=|x|=1$.
(For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)
For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
And it is possible to show that for each $xin X$ there is exactly one functional $f$
with $|f|=f(x)=|x|=1$.
I hope this is sufficient motivation for the term "smooth" for a normed space.
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
answered Nov 28 '18 at 9:09
supinfsupinf
6,5691028
6,5691028
add a comment |
add a comment |
$begingroup$
This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
Indeed, the set
$$
{ fin X^* mid |f| le 1 text{ and } f(x) = |x|}$$
coincides with the convex subdifferential of $F$ at $x$.
If this subdifferential is a singleton, then $F$ is differentiable.
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
$endgroup$
$begingroup$
$F'(x)=f$ is a typo?
$endgroup$
– David C. Ullrich
Nov 28 '18 at 15:41
$begingroup$
No, the derivative of $F$ at $x$ is the linear function $f$.
$endgroup$
– gerw
Nov 28 '18 at 18:01
$begingroup$
The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
$endgroup$
– David C. Ullrich
Nov 28 '18 at 18:31
add a comment |
$begingroup$
This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
Indeed, the set
$$
{ fin X^* mid |f| le 1 text{ and } f(x) = |x|}$$
coincides with the convex subdifferential of $F$ at $x$.
If this subdifferential is a singleton, then $F$ is differentiable.
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
$endgroup$
$begingroup$
$F'(x)=f$ is a typo?
$endgroup$
– David C. Ullrich
Nov 28 '18 at 15:41
$begingroup$
No, the derivative of $F$ at $x$ is the linear function $f$.
$endgroup$
– gerw
Nov 28 '18 at 18:01
$begingroup$
The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
$endgroup$
– David C. Ullrich
Nov 28 '18 at 18:31
add a comment |
$begingroup$
This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
Indeed, the set
$$
{ fin X^* mid |f| le 1 text{ and } f(x) = |x|}$$
coincides with the convex subdifferential of $F$ at $x$.
If this subdifferential is a singleton, then $F$ is differentiable.
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
$endgroup$
This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
Indeed, the set
$$
{ fin X^* mid |f| le 1 text{ and } f(x) = |x|}$$
coincides with the convex subdifferential of $F$ at $x$.
If this subdifferential is a singleton, then $F$ is differentiable.
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
edited Nov 28 '18 at 21:59
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
answered Nov 28 '18 at 10:31
gerwgerw
19.7k11334
19.7k11334
$begingroup$
$F'(x)=f$ is a typo?
$endgroup$
– David C. Ullrich
Nov 28 '18 at 15:41
$begingroup$
No, the derivative of $F$ at $x$ is the linear function $f$.
$endgroup$
– gerw
Nov 28 '18 at 18:01
$begingroup$
The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
$endgroup$
– David C. Ullrich
Nov 28 '18 at 18:31
add a comment |
$begingroup$
$F'(x)=f$ is a typo?
$endgroup$
– David C. Ullrich
Nov 28 '18 at 15:41
$begingroup$
No, the derivative of $F$ at $x$ is the linear function $f$.
$endgroup$
– gerw
Nov 28 '18 at 18:01
$begingroup$
The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
$endgroup$
– David C. Ullrich
Nov 28 '18 at 18:31
$begingroup$
$F'(x)=f$ is a typo?
$endgroup$
– David C. Ullrich
Nov 28 '18 at 15:41
$begingroup$
$F'(x)=f$ is a typo?
$endgroup$
– David C. Ullrich
Nov 28 '18 at 15:41
$begingroup$
No, the derivative of $F$ at $x$ is the linear function $f$.
$endgroup$
– gerw
Nov 28 '18 at 18:01
$begingroup$
No, the derivative of $F$ at $x$ is the linear function $f$.
$endgroup$
– gerw
Nov 28 '18 at 18:01
$begingroup$
The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
$endgroup$
– David C. Ullrich
Nov 28 '18 at 18:31
$begingroup$
The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
$endgroup$
– David C. Ullrich
Nov 28 '18 at 18:31
add a comment |
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