Equivalence between Hölder norms

Question

Let us consider the following two norms: $$ \left\lVert f\right\rVert_\alpha = \left\lVert f\right\rVert_\infty + \displaystyle{\sup_{\substack{x,y \in U \\ x \neq y}} \frac{\left| f(x) - f(y))\right|}{\left| x-y \right|^\alpha}}, $$ $$ \left\lVert f\right\rVert_{\alpha • } = |f(0)|+ \displaystyle{\sup_{\substack{x,y \in U \\ x \neq y}} \frac{\left| f(x) - f(y))\right|}{\left| x-y \right|^\alpha}}. $$ I need to show that those two norms are equivalent on $\mathcal{C}^\alpha(U,\mathbb{R})$, where $U=[0,T]$ that is the space of $\alpha$-Hölder functions from $U$ to $\mathbb{R}$, then I have to find two constants, for instance let us call them $a,b \in \mathbb{R}$, such that: $$ a \left\lVert f\right\rVert_{\alpha • } \leq \left\lVert f\right\rVert_{\alpha } \leq b \left\lVert f\right\rVert_{\alpha • }. $$ I have issues in handling the $\infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?

Answer 1

\begin{eqnarray} |f(x)| &\le& |f(0)| + |f(x)-f(0)| \\ &\le& |f(0)| + \|f\|_{\alpha • }|x-0|^\alpha \\ &\le& \|f\|_{\alpha • } + T^\alpha \|f\|_{\alpha • } \\ &=& (1+T^\alpha)\|f\|_{\alpha • } \end{eqnarray} And so $\|f\|_\infty \le (1+T^\alpha) \|f\|_{\alpha • }$ and then $\|f\|_{\alpha} \le (2+T^\alpha) \|f\|_{\alpha • }$.

The other direction is immediate since $|f(0)| \le \|f\|_\infty$ and so $\|f\|_{\alpha • } \le \|f\|_{\alpha}$.

Answer 2

It is clear that $\|f\|_{\alpha • } \leq \|f\|_\alpha$. Essentially the only difficult part in showing the other inequality is controlling $\|f\|_\infty$ in terms of $\|f\|_{\alpha • }$ so I will do this.

For $x \in [0,T]$ we have \begin{align*} |f(x)| &\leq |f(0)| + |f(x) - f(0)| \\ & \leq |f(0)| + \|f\|_{\alpha • } |x|^\alpha \\& \leq \|f\|_{\alpha • } + \|f\|_{\alpha • } T^\alpha \\& = (1+ T^\alpha) \|f\|_{\alpha • } \end{align*} Taking the $\sup$ over $U$ gives an inequality of the desired form.