We have $||u||_{C^\alpha,\Omega} = \text{sup}_\Omega |u(x)|+ \text{sup}_\Omega \frac{|u(x)-u(y)|}{|x-y|^\alpha}$
and $||u||_{C^1} =\text{sup}_\Omega |u(x)| + \text{sup}_\Omega|\frac{du}{dx}|$
I have tried a few examples and it seems that $||u||_{C^\alpha} \leq ||u||_{C^1}$ is always true. For instance:
1) The function $u(x) = x^n, n \in \mathbb{N}$, with domain $\Omega = [-1,1]$ has $||u||_{C^\alpha} = 1/1^\alpha = 1$ and $||u||_{C^1} = \text{sup}_\Omega |nx| = n \geq 1$.
2) The function $u(x) = \sin(x)$, with domain $\Omega = [-\frac{\pi}{2},\frac{\pi}{2}]$. We have $||u||_{C^\alpha} = \frac{1}{{(\frac{\pi}{2})}^\alpha} \leq 1$ and $||u||_{C^1} =\text{sup}_\Omega |\cos(x)| = 1$.
3) The function $u(x) = e^x$, domain $\Omega = [-1,1]$. We have $||u||_{C^\alpha} = \frac{e - \frac{1}{e}}{2^\alpha}$, and $||u||_{C^1} =\text{sup}_\Omega e^x = e \geq \frac{e - \frac{1}{e}}{2^\alpha}$
Is this always true? How would one prove it?
First show $\mathcal C^1\subset \mathcal C^{0,1}$:
$$f(x)-f(y)=\int_0^1\frac{d}{dt}(y+t(x-y))dt=\int_0^1 f'(y+t(x-y))(x-y)dt$$ and so
$$|f(x)-f(y)|\leq \max_{z\in[x,y]}|f'(z)||x-y|\leq \|f\|_{C^1}|x-y|$$
Then show $\mathcal C^{0,\beta}\subset \mathcal C^{0,\alpha}$ for all $0<\alpha\leq \beta\leq 1$
I'll note $[f]_\gamma=\sup\frac{|f(x)-f(y)|}{|x-y|^\gamma }$. We immediatly have $$[f]_\alpha\leq[f]_{\beta}$$
Moreover, $$\sup_{|x-y|\geq 1}\frac{|f(x)-f(y)|}{|x-y|}\leq \sup{|f(x)-f(y)|}\leq 2\|f\|_{C^0}$$ and so $$\|f\|_{C^{0,\alpha}}=\|f\|_{C^0}+[f]_\alpha\leq\|f\|_{C^0}+\max\{2\|f\|_{C^0},\|f\|_{C^{0,\beta}}\}\leq 3\|f\|_{C^{0,\beta}}$$
We conclude that $\mathcal C^{1}\subset \mathcal C^{0,\alpha}$