Mixing
Holder continuity of the derivate of $|x|^alpha$ for $alpha>1$
up vote
1
down vote
favorite
Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
$$
f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
$$
If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.
We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
$$
frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
$$
if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.
I try proof that exist $C>0$ such that
$$
|f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
$$
but I only succeeded in the case $m=1$.
functional-analysis analysis inequality pde
asked 2 days ago
Raoní Cabral Ponciano
759
add a comment |
up vote
1
down vote
favorite
Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
$$
f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
$$
If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.
We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
$$
frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
$$
if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.
I try proof that exist $C>0$ such that
$$
|f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
$$
but I only succeeded in the case $m=1$.
functional-analysis analysis inequality pde
asked 2 days ago
Raoní Cabral Ponciano
759
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
$$
f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
$$
If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.
We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
$$
frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
$$
if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.
I try proof that exist $C>0$ such that
$$
|f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
$$
but I only succeeded in the case $m=1$.
functional-analysis analysis inequality pde
asked 2 days ago
Raoní Cabral Ponciano
759
Suppose $U=B(0,1)$ is an open ball in $mathbb R^m$ and $alpha>1$. My question is if $|x|^alphain C^{1,gamma}(U)$ for some $gammain (0,1]$. I know that $|x|^alphain C^1(U)$ and
$$
f_i(x):=frac{partial}{partial x_i}|x|^alpha=alpha|x|^{alpha-2}x_i, mbox{ if }xneq 0.
$$
If $alphageq 2$, it is easy because we obtain $|x|^alphain C^2(U)subset C^{1,1}(U)$, with $gamma=1$ in this case.
We can suppose $1<alpha<2$. If $|x|^alphain C^{1,gamma}(U)$ then using $x_n=frac{e_i}n$ and $y_n=-frac{e_i}n$ we obtain
$$
frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^gamma}=frac{alpha}{2^{gamma-1}}n^{gamma-(alpha-1)}overset{nrightarrow+infty}longrightarrow+infty,
$$
if $gamma>alpha-1$. If this $gamma$ exist we must have $gammaleqalpha-1$. I guess that $|x|^alphain C^{1,alpha-1}(U)$.
I try proof that exist $C>0$ such that
$$
|f_i(x)-f_i(y)|leq C|x-y|^{alpha-1},
$$
but I only succeeded in the case $m=1$.
functional-analysis analysis inequality pde
functional-analysis analysis inequality pde
asked 2 days ago
Raoní Cabral Ponciano
759
asked 2 days ago
Raoní Cabral Ponciano
759
asked 2 days ago
Raoní Cabral Ponciano
759
asked 2 days ago
Raoní Cabral Ponciano
759
asked 2 days ago
Raoní Cabral Ponciano
759
759
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
First, let me prove this for $m=1$. The question is, whether the function
$$
g(x):= sign(x)|x|^gamma
$$
is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
$$
g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
$$
and by Hoelder's inequality
$$
|g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
= gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
le gamma (x-y)^gamma .
$$
In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
$$
frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
$$
This shows $gin C^gamma([-1,1])$.
Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
$$
f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
$$
The second term can be reduced to the case $m=1$. The first term is unproblematic
if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
Then
$$
left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
$$
This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .
answered 2 days ago
daw
23.8k1544
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First, let me prove this for $m=1$. The question is, whether the function
$$
g(x):= sign(x)|x|^gamma
$$
is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
$$
g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
$$
and by Hoelder's inequality
$$
|g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
= gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
le gamma (x-y)^gamma .
$$
In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
$$
frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
$$
This shows $gin C^gamma([-1,1])$.
Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
$$
f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
$$
The second term can be reduced to the case $m=1$. The first term is unproblematic
if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
Then
$$
left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
$$
This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .
answered 2 days ago
daw
23.8k1544
add a comment |
up vote
1
down vote
accepted
First, let me prove this for $m=1$. The question is, whether the function
$$
g(x):= sign(x)|x|^gamma
$$
is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
$$
g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
$$
and by Hoelder's inequality
$$
|g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
= gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
le gamma (x-y)^gamma .
$$
In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
$$
frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
$$
This shows $gin C^gamma([-1,1])$.
Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
$$
f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
$$
The second term can be reduced to the case $m=1$. The first term is unproblematic
if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
Then
$$
left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
$$
This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .
answered 2 days ago
daw
23.8k1544
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First, let me prove this for $m=1$. The question is, whether the function
$$
g(x):= sign(x)|x|^gamma
$$
is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
$$
g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
$$
and by Hoelder's inequality
$$
|g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
= gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
le gamma (x-y)^gamma .
$$
In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
$$
frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
$$
This shows $gin C^gamma([-1,1])$.
Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
$$
f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
$$
The second term can be reduced to the case $m=1$. The first term is unproblematic
if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
Then
$$
left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
$$
This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .
answered 2 days ago
daw
23.8k1544
First, let me prove this for $m=1$. The question is, whether the function
$$
g(x):= sign(x)|x|^gamma
$$
is in $C^gamma([-1,1])$ for $gammain (0,1)$. Let $x,y$ be such that $0<x<y$. Then
$$
g(y)-g(x) = int_x^y gamma t^{gamma-1}dt
$$
and by Hoelder's inequality
$$
|g(y)-g(x)| le int_x^y gamma t^{gamma-1}dt le gamma (x-y)^gamma cdot left(int_x^y |t|^{(gamma-1)cdot frac{gamma}{1-gamma}} dtright)^{frac{1-gamma}gamma}
= gamma (x-y)^gamma ( y^{1-gamma}-x^{1-gamma})^{frac{1-gamma}gamma}
le gamma (x-y)^gamma .
$$
In case $0<x<y$, we have $|x-y| ge |x|,|y|$, which implies
$$
frac{g(y)-g(x)}{|y-x|^gamma} =frac{|y|^gamma+ |x|^gamma}{|y-x|^gamma} le2.
$$
This shows $gin C^gamma([-1,1])$.
Let now $m>1$. Set $gamma:=alpha-1$. Then with your functions $f_i$ we have for $0<|x|le |y|$
$$
f_i(x)- f_i(y) = gamma( |x|^gamma ( frac{x_i}{|x|} - frac{y_i}{|y|})+ (|x|^{gamma-1} - |y|^{gamma-1}) frac{y_i}{|y|}).
$$
The second term can be reduced to the case $m=1$. The first term is unproblematic
if $|x|le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|le 2|x|$.
Then
$$
left|frac{x_i}{|x|} - frac{y_i}{|y|} right|=left| frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|cdot |y|}right|lefrac{(|x|+|y|)|x-y|}{|x|cdot |y|}
le 2frac{|x-y|}{|x|} le 2|x-y|^gamma |x|^{-gamma}.
$$
This enables to prove $f_iin C^gamma(overline{B_1(0)})$ and $fin C^alpha(overline{B_1(0)})$ .
answered 2 days ago
daw
23.8k1544
answered 2 days ago
daw
23.8k1544
answered 2 days ago
daw
23.8k1544
answered 2 days ago
daw
23.8k1544
23.8k1544
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005530%2fholder-continuity-of-the-derivate-of-x-alpha-for-alpha1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
