Hölder continuous functions from $L^1(\mathbb{R})$

Question

Let $f\in L^1(\mathbb{R})$ be a Hölder continuous function of order $\alpha$.

Can we always find constants $\varepsilon>0$ and $C>0$ such that $$|f(x)-f(y)|\leq C\frac{|x-y|^\alpha}{(1+|x|+|y|)^{1+\varepsilon}}, \quad \textrm{for all } x,y\in\mathbb{R}?$$

Answer 1

No (unless $f=0$). Fix any $y_0$ with $|f(y_0)|>0$. Then $$ |f(y_0)|-|f(x)| \le |f(x)-f(y_0)| \le C\frac{|x-y_0|^\alpha}{(1+|x|+|y_0|)^{1+\varepsilon}} $$ $$ \Rightarrow |f(y_0)|- C\frac{|x-y_0|^\alpha}{(1+|x|+|y_0|)^{1+\varepsilon}} \le |f(x)| $$ As $\alpha < 1+ \varepsilon$ the left hand side in this inequality tends to $|f(y_0)|$ as $|x| \to \infty$. But then $f \in L^1$ is not possible.