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Closure of Continuously DifferentiableFunctions in Holder Space

Clarification of Hölder norm in terms of oscillationfunctions in Holder space$C^1$ is not dense in Hölder spaceHolder Continuous Functions on $[0,1]$ are complete + Banach spaceDual of the Banach space of $k$-times continuously differentiable functions.Can $C^k(overlineOmega)$ functions extend to $C^k(mathbbR^n)$?denseness of smooth functions in space of lipschitz continuous functionsPointwise convergence in Holder spaceContinuity of the differential operator from $(C^1[0,1],d_infty)$ to $(C[0,1],d_1)$Compact Embedding Between Parabolic Holder Spaces

0

$begingroup$

Here is a curious (already submitted) homework problem I had in analysis some time ago:

Let $Omega$ be a convex domain in $mathbbR^n$ with $C^1$ boundary. Let $C^0,alpha(overlineOmega)$ denote those functions that are Holder continuous on $overlineOmega$ with coefficient $alpha$ and with finite Holder norm:
$$
|f|_alpha = |f|_infty + sup_x neq yfracx-y
$$
Let $C^1(overlineOmega)$ denote the continuously differentiable functions on $overlineOmega$. Show that the continuously differentiable functions are not dense in $C^0,alpha$ under the Holder norm, and moreover that a function $f in C^0,alpha$ lies in the closure of $C^1$ (w.r.t. Holder Norm) if and only if for any $x in overlineOmega$:

$$
lim_(y,z) to (x,x)fracy-z = 0
$$
A hint is provided, stating that because $overlineOmega$ is closed + convex, we have a projection $eta(x)$ that allows us to extend $f$ to all of $mathbbR^n$, and moreover, the projection is Lipschitz ($|eta(x) - eta(y)| leq|x-y|$) thanks to projections being distance non-increasing.

real-analysis functional-analysis pde holder-spaces

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edited Mar 13 at 17:34

rubikscube09

asked Mar 13 at 17:21

rubikscube09rubikscube09

1,563720

$endgroup$

add a comment | 

0

$begingroup$

Here is a curious (already submitted) homework problem I had in analysis some time ago:

Let $Omega$ be a convex domain in $mathbbR^n$ with $C^1$ boundary. Let $C^0,alpha(overlineOmega)$ denote those functions that are Holder continuous on $overlineOmega$ with coefficient $alpha$ and with finite Holder norm:
$$
|f|_alpha = |f|_infty + sup_x neq yfracx-y
$$
Let $C^1(overlineOmega)$ denote the continuously differentiable functions on $overlineOmega$. Show that the continuously differentiable functions are not dense in $C^0,alpha$ under the Holder norm, and moreover that a function $f in C^0,alpha$ lies in the closure of $C^1$ (w.r.t. Holder Norm) if and only if for any $x in overlineOmega$:

$$
lim_(y,z) to (x,x)fracy-z = 0
$$
A hint is provided, stating that because $overlineOmega$ is closed + convex, we have a projection $eta(x)$ that allows us to extend $f$ to all of $mathbbR^n$, and moreover, the projection is Lipschitz ($|eta(x) - eta(y)| leq|x-y|$) thanks to projections being distance non-increasing.

real-analysis functional-analysis pde holder-spaces

share|cite|improve this question

edited Mar 13 at 17:34

rubikscube09

asked Mar 13 at 17:21

rubikscube09rubikscube09

1,563720

$endgroup$

add a comment | 

$begingroup$

Here is a curious (already submitted) homework problem I had in analysis some time ago:

Let $Omega$ be a convex domain in $mathbbR^n$ with $C^1$ boundary. Let $C^0,alpha(overlineOmega)$ denote those functions that are Holder continuous on $overlineOmega$ with coefficient $alpha$ and with finite Holder norm:
$$
|f|_alpha = |f|_infty + sup_x neq yfracx-y
$$
Let $C^1(overlineOmega)$ denote the continuously differentiable functions on $overlineOmega$. Show that the continuously differentiable functions are not dense in $C^0,alpha$ under the Holder norm, and moreover that a function $f in C^0,alpha$ lies in the closure of $C^1$ (w.r.t. Holder Norm) if and only if for any $x in overlineOmega$:

$$
lim_(y,z) to (x,x)fracy-z = 0
$$
A hint is provided, stating that because $overlineOmega$ is closed + convex, we have a projection $eta(x)$ that allows us to extend $f$ to all of $mathbbR^n$, and moreover, the projection is Lipschitz ($|eta(x) - eta(y)| leq|x-y|$) thanks to projections being distance non-increasing.

real-analysis functional-analysis pde holder-spaces

share|cite|improve this question

edited Mar 13 at 17:34

rubikscube09

asked Mar 13 at 17:21

rubikscube09rubikscube09

1,563720

$endgroup$

share|cite|improve this question

edited Mar 13 at 17:34

rubikscube09

asked Mar 13 at 17:21

rubikscube09rubikscube09

1,563720

share|cite|improve this question

edited Mar 13 at 17:34

rubikscube09

asked Mar 13 at 17:21

rubikscube09rubikscube09

1,563720

asked Mar 13 at 17:21

rubikscube09rubikscube09

1,563720

asked Mar 13 at 17:21

rubikscube09rubikscube09

1,563720

1 Answer
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oldest

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1

$begingroup$

It would appear that $C^1$ is not dense in $C^0,alpha$ for any $Omega.$ To start, take $n=1$ and $Omega = (-1,1).$ Fix $alpha in (0,1).$ Let $f(x)=|x|^alpha.$ Then $f in C^0,alpha(-1,1).$ Suppose $gin C^1(-1,1).$ We have

$$|f-g|_alpha ge frac $$ $$= fracx^alpha ge 1- fracg(x)-g(0)^alpha$$

for $x$ close to $0.$ As $xto 0,$ the last fraction $to 0$ since $g'(0)$ exists. It follows that $|f-g|_alpha ge 1.$

This idea should work for any $n$ by taking $f(x_1,dots ,x_n)=|x_1|^alpha.$

share|cite|improve this answer

answered Mar 13 at 19:41

zhw.zhw.

74.5k43175

$endgroup$

add a comment | 

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1

$begingroup$

It would appear that $C^1$ is not dense in $C^0,alpha$ for any $Omega.$ To start, take $n=1$ and $Omega = (-1,1).$ Fix $alpha in (0,1).$ Let $f(x)=|x|^alpha.$ Then $f in C^0,alpha(-1,1).$ Suppose $gin C^1(-1,1).$ We have

$$|f-g|_alpha ge frac $$ $$= fracx^alpha ge 1- fracg(x)-g(0)^alpha$$

for $x$ close to $0.$ As $xto 0,$ the last fraction $to 0$ since $g'(0)$ exists. It follows that $|f-g|_alpha ge 1.$

This idea should work for any $n$ by taking $f(x_1,dots ,x_n)=|x_1|^alpha.$

share|cite|improve this answer

answered Mar 13 at 19:41

zhw.zhw.

74.5k43175

$endgroup$

add a comment | 

1

$begingroup$

It would appear that $C^1$ is not dense in $C^0,alpha$ for any $Omega.$ To start, take $n=1$ and $Omega = (-1,1).$ Fix $alpha in (0,1).$ Let $f(x)=|x|^alpha.$ Then $f in C^0,alpha(-1,1).$ Suppose $gin C^1(-1,1).$ We have

$$|f-g|_alpha ge frac $$ $$= fracx^alpha ge 1- fracg(x)-g(0)^alpha$$

for $x$ close to $0.$ As $xto 0,$ the last fraction $to 0$ since $g'(0)$ exists. It follows that $|f-g|_alpha ge 1.$

This idea should work for any $n$ by taking $f(x_1,dots ,x_n)=|x_1|^alpha.$

share|cite|improve this answer

answered Mar 13 at 19:41

zhw.zhw.

74.5k43175

$endgroup$

add a comment | 

$begingroup$

It would appear that $C^1$ is not dense in $C^0,alpha$ for any $Omega.$ To start, take $n=1$ and $Omega = (-1,1).$ Fix $alpha in (0,1).$ Let $f(x)=|x|^alpha.$ Then $f in C^0,alpha(-1,1).$ Suppose $gin C^1(-1,1).$ We have

$$|f-g|_alpha ge frac $$ $$= fracx^alpha ge 1- fracg(x)-g(0)^alpha$$

for $x$ close to $0.$ As $xto 0,$ the last fraction $to 0$ since $g'(0)$ exists. It follows that $|f-g|_alpha ge 1.$

This idea should work for any $n$ by taking $f(x_1,dots ,x_n)=|x_1|^alpha.$

share|cite|improve this answer

answered Mar 13 at 19:41

zhw.zhw.

74.5k43175

$endgroup$

share|cite|improve this answer

answered Mar 13 at 19:41

zhw.zhw.

74.5k43175

answered Mar 13 at 19:41

zhw.zhw.

74.5k43175

answered Mar 13 at 19:41

zhw.zhw.

74.5k43175