Show: $C^1(\Omega)\subset C^{0,1}(\Omega)\subset C^{0,\lambda}(\Omega)\subset C^0(\Omega)$.

Question

Show that $$ C^1(\Omega)\subset C^{0,1}(\Omega)\subset C^{0,\lambda}(\Omega)\subset C^0(\Omega)~~~~~~~\forall0<\lambda\leq 1. $$

Hello, I have some problems to show these inclusions! In order to get some help, I wrote down all the definitions that we had (please see below).

So consider $u\in C^1(\Omega)$. Concerning to the second definition of 1.), then $u$ is continious on the domain $\Omega$ and the first partial derivatives are continious on $\Omega$.

Now I have to show, that $u\in C^{0,1}(\Omega)$.

Which definition do I have to use resp. what do I have to show?

---

This are the definitions that we wrote down: (unfortunately rather many)

1.) $\Omega\subset\mathbb{R}^n$ a domain and $k\in\mathbb{N}_0$

$$ C(\Omega):=C^0(\Omega):=\left\{u\colon\Omega\to\mathbb{R}|u\mbox{ continious on }\Omega\right\} $$

$$ C^k(\Omega):=\left\{u\in C(\Omega)|\forall\alpha\in\mathbb{N}_0^n\mbox{ with }0\leq\lvert\alpha\rvert\leq k: D^{\alpha} u(x)\in C(\Omega)\right\} $$

$$ C^{\infty}(\Omega):=\bigcap\limits_{k=0}^{\infty}C^k(\Omega) $$

2.) $\Omega\subset\mathbb{R}^n$ limited domain and $k\in\mathbb{N}_0$

$$ C^k(\overline{\Omega}):=\left\{u\in C^k(\Omega)|\forall\alpha\in\mathbb{N}_0^{n}\mbox{ with }\lvert\alpha\rvert\leq k\mbox{ is }D^{\alpha}u\mbox{ uniformly continious on }\Omega\right\} $$

3.) $\Omega\subset\mathbb{R}^n$ a (not necessarily limited)) domain and $k\in\mathbb{N}_0$

$$ C^k(\overline{\Omega}):=\left\{u\colon\overline{\Omega}\to\mathbb{R}|\forall~R>0: u\in C^k(\overline{B_R(0)\cap\Omega})\right\} $$

4.) $\Omega\subset\mathbb{R}^n$ limited domain, $\lambda\in ]0,1]$ and $k\in\mathbb{N}_0$

$$ C^{k,\lambda}(\overline{\Omega}):=\left\{u\in C^k(\overline{\Omega})|\forall\alpha\in\mathbb{N}_0^n\mbox{ with }\lvert\alpha\rvert\leq k: H_{\alpha,\lambda}(u)<\infty\right\}, $$

whereat

$$ H_{\alpha,\lambda}(u):=\sup\limits_{x,y\in\overline{\Omega}, x\neq y}\left\{\frac{\lvert D^{\alpha} u(x)-D^{\alpha} u(y)\rvert}{\lVert x-y\rVert^{\lambda}}\right\} $$

5.) $\Omega\subset\mathbb{R}^n$ a (not necessarily limited) domain, $\lambda\in ]0,1]$ and $k\in\mathbb{N}_0$

$$ C^{k,\lambda}(\Omega):=\left\{u\in C^k(\Omega)|\forall\overline{B}_R(x)\subset\Omega: u\in C^{k,\lambda}(\overline{B}_R(x))\right\} $$

$$ C^{k,\lambda}(\overline{\Omega}):=\left\{u\colon\overline{\Omega}\to\mathbb{R}|\forall R>0: u\in C^{k,\lambda}(\overline{B_R(0)\cap\Omega})\right\} $$

Answer 1

I - $C^1(\Omega)\subset C^{0,1}(\Omega)$

Take any $B_R(x)$ such that $\overline{B_R(x)}\subset \Omega$. If $u\in C^1(\Omega)$ then $u\in C^1(\overline{B_R(x)})$, which implies that $\|u'(x)\|\leq C$, where $C>0$ is a constant.

Can you concude from here?

Hint: Apply Mean Value Theorem.

II - $C^{0,1}(\Omega)\subset C^{0,\lambda}(\Omega)$

Take any $B_R(x)$ such that $\overline{B_R(x)}\subset \Omega$ and $z,y\in \overline{B_R(x)}$. Note

\begin{eqnarray} \frac{\|u(z)-u(y)\|}{\|z-y\|^\lambda} &=& \frac{\|u(z)-u(y)\|}{\|z-y\|}\frac{\|z-y\|}{\|z-y\|^\lambda} \nonumber \\ &=& \frac{\|u(z)-u(y)\|}{\|z-y\|}\|z-y\|^{1-\lambda}\nonumber \end{eqnarray}

Can you conclude from here?

III - $C^{0,\lambda}(\Omega)\subset C^0(\Omega)$

This is immediate from definition.