Can someone come up with a proof for this bound of the Fourier transformation?
Let $f$ be $s$ times continuously differentiable, with compact support, then there exists a real constant $c$ such that:
$$\vert F(k)\vert\leq \frac{c}{(1+\vert k\vert)^s},$$ where $F$ is the Fourier transform of $f$.
Since there is a compact support, the maximum of the function exists and the Fourier transformation is bounded, but where does the factor $(1 +\vert k\vert)^{-s}$ come from?
Thanks
So first a correction, it is not because a function is bounded that the Fourier transform will be bounded (take for example $f=1$. Then its Fourier transform is the dirac distribution).
Now about your inequality, the rule to remember is that the decay of the Fourier transform is linked to the regularity of the function because $\mathcal{F}(\nabla f) = -i\,y \hat{f}(y)$ (where I denote by $\hat{f} = \mathcal{F}(f)$ the Fourier transform of $f$. On the other hand, since it is an integral transform, the Fourier transform of an $L^1$ function will be bounded. Mixing this two ideas yields $$ \begin{align*} (1+|y|^s)\,|\hat{f}(y)| &= |\hat{f}(y)| + |y^s\hat{f}(y)| \\ &= |\hat{f}(y)| + |\nabla^s\hat{f}(y)| \\ &≤ ∫_{\mathbb{R^d}}|f(x)|\,\mathrm{d} x + ∫_{\mathbb{R^d}}|\nabla^sf(x)|\,\mathrm{d} x. \end{align*} $$ Since in your particular case, your function is compactly supported, say on a bounded set $\Omega$, the integrals here are bounded by $$ \begin{align*} ∫_{\mathbb{R^d}}|f(x)|\,\mathrm{d} x + ∫_{\mathbb{R^d}}|\nabla^sf(x)|\,\mathrm{d} x &= ∫_{\Omega}|f(x)|\,\mathrm{d} x + ∫_{\Omega}|\nabla^sf(x)|\,\mathrm{d} x \\ &\leq |\Omega|\left(\|f\|_{L^\infty} + \|\nabla^sf\|_{L^\infty}\right) =: c_0 \end{align*} $$ And so $$ |\hat{f}(y)| ≤ \frac{c_0}{1+|y|^s} $$
If you want it exactly on the form you have, just remark that the convexity of $x↦|x|^s$ $$ (1+|y|)^s ≤ 2^{s-1}\left(1+|y|^s\right) $$ so that $$ |\hat{f}(y)| ≤ \frac{c}{(1+|y|)^s} $$ with $c = c_0\, 2^{s-1}$.