Mixing
Differentiability at the origin of Norm
$begingroup$
For $alpha > 1$, show that $f:mathbb{R}^n to mathbb{R}^n$, given by
$$f(x)=rVert xrVert^alpha$$
is differentiable at the origin.
What I did:
What is necessary is to find a linear transformation such that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=0$. $f(h)-f(0) = rVert hrVert ^alpha$. So, because $alpha>1$, I guess that T=0 so that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=lim_{hto 0}frac{rVert hrVert ^alpha}{rVert h rVert}=lim_{hto 0}rVert hrVert^{alpha-1}=0$.
Is this correct?
real-analysis proof-verification frechet-derivative
edited Jan 28 at 22:08
Scientifica
6,82941335
asked Jan 28 at 21:58
FhoenixFhoenix
14811
$endgroup$
add a comment |
$begingroup$
For $alpha > 1$, show that $f:mathbb{R}^n to mathbb{R}^n$, given by
$$f(x)=rVert xrVert^alpha$$
is differentiable at the origin.
What I did:
What is necessary is to find a linear transformation such that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=0$. $f(h)-f(0) = rVert hrVert ^alpha$. So, because $alpha>1$, I guess that T=0 so that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=lim_{hto 0}frac{rVert hrVert ^alpha}{rVert h rVert}=lim_{hto 0}rVert hrVert^{alpha-1}=0$.
Is this correct?
real-analysis proof-verification frechet-derivative
edited Jan 28 at 22:08
Scientifica
6,82941335
asked Jan 28 at 21:58
FhoenixFhoenix
14811
$endgroup$
2
$begingroup$
Yes, this is correct. However, note that $fcolonBbb R^ntoBbb R$.
$endgroup$
– Ted Shifrin
Jan 28 at 22:07
1
$begingroup$
It is correct. You could also do that without guessing by calculating the partial derivatives. (which is of course not enough to prove differentiability, but that would give you the only linear transformation that might be the derivative)
$endgroup$
– Mark
Jan 28 at 22:11
1
$begingroup$
@Mark if the partial derivatives are continuous this is enough. Moreover how taking the partial derivatives gives you information here about the differential since we don't know the norm he is using ?
$endgroup$
– Thinking
Jan 28 at 22:21
1
$begingroup$
@Thinking, if they are continuous then yes, it is enough. Not so easy to prove they are continuous though. And yes, you are right that OP didn't write what norm it is. I supposed it is the Euclidean norm.
$endgroup$
– Mark
Jan 28 at 22:51
add a comment |
$begingroup$
For $alpha > 1$, show that $f:mathbb{R}^n to mathbb{R}^n$, given by
$$f(x)=rVert xrVert^alpha$$
is differentiable at the origin.
What I did:
What is necessary is to find a linear transformation such that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=0$. $f(h)-f(0) = rVert hrVert ^alpha$. So, because $alpha>1$, I guess that T=0 so that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=lim_{hto 0}frac{rVert hrVert ^alpha}{rVert h rVert}=lim_{hto 0}rVert hrVert^{alpha-1}=0$.
Is this correct?
real-analysis proof-verification frechet-derivative
edited Jan 28 at 22:08
Scientifica
6,82941335
asked Jan 28 at 21:58
FhoenixFhoenix
14811
$endgroup$
For $alpha > 1$, show that $f:mathbb{R}^n to mathbb{R}^n$, given by
$$f(x)=rVert xrVert^alpha$$
is differentiable at the origin.
What I did:
What is necessary is to find a linear transformation such that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=0$. $f(h)-f(0) = rVert hrVert ^alpha$. So, because $alpha>1$, I guess that T=0 so that $lim_{hto 0}frac{f(h)-f(0)-Th}{rVert h rVert}=lim_{hto 0}frac{rVert hrVert ^alpha}{rVert h rVert}=lim_{hto 0}rVert hrVert^{alpha-1}=0$.
Is this correct?
real-analysis proof-verification frechet-derivative
real-analysis proof-verification frechet-derivative
edited Jan 28 at 22:08
Scientifica
6,82941335
asked Jan 28 at 21:58
FhoenixFhoenix
14811
edited Jan 28 at 22:08
Scientifica
6,82941335
asked Jan 28 at 21:58
FhoenixFhoenix
14811
edited Jan 28 at 22:08
Scientifica
6,82941335
edited Jan 28 at 22:08
Scientifica
6,82941335
edited Jan 28 at 22:08
Scientifica
6,82941335
6,82941335
asked Jan 28 at 21:58
FhoenixFhoenix
14811
asked Jan 28 at 21:58
FhoenixFhoenix
14811
asked Jan 28 at 21:58
FhoenixFhoenix
14811
14811
2
$begingroup$
Yes, this is correct. However, note that $fcolonBbb R^ntoBbb R$.
$endgroup$
– Ted Shifrin
Jan 28 at 22:07
1
$begingroup$
It is correct. You could also do that without guessing by calculating the partial derivatives. (which is of course not enough to prove differentiability, but that would give you the only linear transformation that might be the derivative)
$endgroup$
– Mark
Jan 28 at 22:11
1
$begingroup$
@Mark if the partial derivatives are continuous this is enough. Moreover how taking the partial derivatives gives you information here about the differential since we don't know the norm he is using ?
$endgroup$
– Thinking
Jan 28 at 22:21
1
$begingroup$
@Thinking, if they are continuous then yes, it is enough. Not so easy to prove they are continuous though. And yes, you are right that OP didn't write what norm it is. I supposed it is the Euclidean norm.
$endgroup$
– Mark
Jan 28 at 22:51
add a comment |
2
$begingroup$
Yes, this is correct. However, note that $fcolonBbb R^ntoBbb R$.
$endgroup$
– Ted Shifrin
Jan 28 at 22:07
1
$begingroup$
It is correct. You could also do that without guessing by calculating the partial derivatives. (which is of course not enough to prove differentiability, but that would give you the only linear transformation that might be the derivative)
$endgroup$
– Mark
Jan 28 at 22:11
1
$begingroup$
@Mark if the partial derivatives are continuous this is enough. Moreover how taking the partial derivatives gives you information here about the differential since we don't know the norm he is using ?
$endgroup$
– Thinking
Jan 28 at 22:21
1
$begingroup$
@Thinking, if they are continuous then yes, it is enough. Not so easy to prove they are continuous though. And yes, you are right that OP didn't write what norm it is. I supposed it is the Euclidean norm.
$endgroup$
– Mark
Jan 28 at 22:51
2
2
$begingroup$
Yes, this is correct. However, note that $fcolonBbb R^ntoBbb R$.
$endgroup$
– Ted Shifrin
Jan 28 at 22:07
$begingroup$
Yes, this is correct. However, note that $fcolonBbb R^ntoBbb R$.
$endgroup$
– Ted Shifrin
Jan 28 at 22:07
1
1
$begingroup$
It is correct. You could also do that without guessing by calculating the partial derivatives. (which is of course not enough to prove differentiability, but that would give you the only linear transformation that might be the derivative)
$endgroup$
– Mark
Jan 28 at 22:11
$begingroup$
It is correct. You could also do that without guessing by calculating the partial derivatives. (which is of course not enough to prove differentiability, but that would give you the only linear transformation that might be the derivative)
$endgroup$
– Mark
Jan 28 at 22:11
1
1
$begingroup$
@Mark if the partial derivatives are continuous this is enough. Moreover how taking the partial derivatives gives you information here about the differential since we don't know the norm he is using ?
$endgroup$
– Thinking
Jan 28 at 22:21
$begingroup$
@Mark if the partial derivatives are continuous this is enough. Moreover how taking the partial derivatives gives you information here about the differential since we don't know the norm he is using ?
$endgroup$
– Thinking
Jan 28 at 22:21
1
1
$begingroup$
@Thinking, if they are continuous then yes, it is enough. Not so easy to prove they are continuous though. And yes, you are right that OP didn't write what norm it is. I supposed it is the Euclidean norm.
$endgroup$
– Mark
Jan 28 at 22:51
$begingroup$
@Thinking, if they are continuous then yes, it is enough. Not so easy to prove they are continuous though. And yes, you are right that OP didn't write what norm it is. I supposed it is the Euclidean norm.
$endgroup$
– Mark
Jan 28 at 22:51
add a comment |
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2
$begingroup$
Yes, this is correct. However, note that $fcolonBbb R^ntoBbb R$.
$endgroup$
– Ted Shifrin
Jan 28 at 22:07
1
$begingroup$
It is correct. You could also do that without guessing by calculating the partial derivatives. (which is of course not enough to prove differentiability, but that would give you the only linear transformation that might be the derivative)
$endgroup$
– Mark
Jan 28 at 22:11
1
$begingroup$
@Mark if the partial derivatives are continuous this is enough. Moreover how taking the partial derivatives gives you information here about the differential since we don't know the norm he is using ?
$endgroup$
– Thinking
Jan 28 at 22:21
1
$begingroup$
@Thinking, if they are continuous then yes, it is enough. Not so easy to prove they are continuous though. And yes, you are right that OP didn't write what norm it is. I supposed it is the Euclidean norm.
$endgroup$
– Mark
Jan 28 at 22:51