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How to conclude from a homogeneous function $f$ of degree $1$ differentiable at $0$ that all norms are never...
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Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.
How do we conclude from this that all norms are never differentiable at the origin?
functional-analysis differential-geometry
edited Jan 8 at 19:45
mechanodroid
27.3k62447
asked Jan 8 at 17:54
metalder9metalder9
647
$endgroup$
add a comment |
$begingroup$
Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.
How do we conclude from this that all norms are never differentiable at the origin?
functional-analysis differential-geometry
edited Jan 8 at 19:45
mechanodroid
27.3k62447
asked Jan 8 at 17:54
metalder9metalder9
647
$endgroup$
2
$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00
$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32
add a comment |
$begingroup$
Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.
How do we conclude from this that all norms are never differentiable at the origin?
functional-analysis differential-geometry
edited Jan 8 at 19:45
mechanodroid
27.3k62447
asked Jan 8 at 17:54
metalder9metalder9
647
$endgroup$
Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.
How do we conclude from this that all norms are never differentiable at the origin?
functional-analysis differential-geometry
functional-analysis differential-geometry
edited Jan 8 at 19:45
mechanodroid
27.3k62447
asked Jan 8 at 17:54
metalder9metalder9
647
edited Jan 8 at 19:45
mechanodroid
27.3k62447
asked Jan 8 at 17:54
metalder9metalder9
647
edited Jan 8 at 19:45
mechanodroid
27.3k62447
edited Jan 8 at 19:45
mechanodroid
27.3k62447
edited Jan 8 at 19:45
mechanodroid
27.3k62447
27.3k62447
asked Jan 8 at 17:54
metalder9metalder9
647
asked Jan 8 at 17:54
metalder9metalder9
647
asked Jan 8 at 17:54
metalder9metalder9
647
647
2
$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00
$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32
add a comment |
2
$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00
$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32
2
2
$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00
$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00
$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32
$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32
add a comment |
1 Answer
1
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oldest
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To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
$$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$
Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.
Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
$$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$
so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.
The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
$endgroup$
add a comment |
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1 Answer
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$begingroup$
To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
$$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$
Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.
Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
$$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$
so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.
The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
$endgroup$
add a comment |
$begingroup$
To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
$$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$
Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.
Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
$$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$
so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.
The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
$endgroup$
add a comment |
$begingroup$
To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
$$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$
Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.
Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
$$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$
so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.
The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
$endgroup$
To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
$$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$
Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.
Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
$$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$
so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.
The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
answered Jan 8 at 19:43
mechanodroidmechanodroid
27.3k62447
27.3k62447
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2
$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00
$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32