Holder continuity of the derivate of $|x|^alpha$ for $alpha>1$
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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 Nov 19 at 21:12
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 Nov 19 at 21:12
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 Nov 19 at 21:12
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 Nov 19 at 21:12
Raoní Cabral Ponciano
759
asked Nov 19 at 21:12
Raoní Cabral Ponciano
759
asked Nov 19 at 21:12
Raoní Cabral Ponciano
759
asked Nov 19 at 21:12
Raoní Cabral Ponciano
759
asked Nov 19 at 21:12
Raoní Cabral Ponciano
759
759
add a comment |
add a comment |
1 Answer
1
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oldest
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up vote
1
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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 Nov 20 at 8:03
daw
24k1544
add a comment |
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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 Nov 20 at 8:03
daw
24k1544
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 Nov 20 at 8:03
daw
24k1544
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 Nov 20 at 8:03
daw
24k1544
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 Nov 20 at 8:03
daw
24k1544
answered Nov 20 at 8:03
daw
24k1544
answered Nov 20 at 8:03
daw
24k1544
answered Nov 20 at 8:03
daw
24k1544
24k1544
add a comment |
add a comment |
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