Suppose that $f$ is continuous on $[0,1]$. Thus, $f\in L^1(\mathbb{R})$ and its forier transform exists, as $$ \hat{f}(\xi) := \int_\mathbb{R} e^{-2\pi i x \xi} f(x)dx, $$ which can also be written as $$ \hat{f}(\xi) = -\int_\mathbb{R} e^{-2\pi i \xi (x-\frac{1}{2\xi})} f(x)dx = \int_\mathbb{R} e^{-2\pi i x \xi} (f(x)-f(x+\frac{1}{2\xi}))dx.$$ Therefore, if $f$ satisfies an Holder condition $f(x)-f(x+h)=O(h^a)$ for small $h$, where $O$ is independent to $x$, it holds $\hat{f}(\xi) = O(\xi^{-a})$ for large $\xi$.

My question is whether we may obtain $f$ satisfies some Holder condition if we know that $\hat{f}(\xi) = O(\xi^{-a})$ for large $\xi$. If not, what other condition should be restricted to $f$ or $\hat{f}$ to get the result?

Suppose that $f$ is continuous on $[0,1]$. Thus, $f\in L^1(\mathbb{R})$ and its Fourier transform exists, as $$ \hat{f}(\xi) := \int_\mathbb{R} e^{-2\pi i x \xi} f(x)dx, $$ which can also be written as $$ \hat{f}(\xi) = -\int_\mathbb{R} e^{-2\pi i \xi (x-\frac{1}{2\xi})} f(x)dx = \int_\mathbb{R} e^{-2\pi i x \xi} (f(x)-f(x+\frac{1}{2\xi}))dx.$$ Therefore, if $f$ satisfies a Hölder condition $f(x)-f(x+h)=O(h^a)$ for small $h$, where $O$ is independent to $x$, it holds $\hat{f}(\xi) = O(\xi^{-a})$ for large $\xi$.

My question is whether we may obtain $f$ satisfies some Hölder condition if we know that $\hat{f}(\xi) = O(\xi^{-a})$ for large $\xi$. If not, what other condition should be restricted to $f$ or $\hat{f}$ to get the result?