Show that $\hat{f}$ is continuous and $\hat{f(\xi)}\rightarrow 0 $ as $|\xi|\rightarrow 0$ Suppose $f$ continuous and of moderate decrease. Denote the Fourier Transform of $f$ by $\hat{f}$. I need to prove that $\hat{f}$ is continuous and $\hat{f(\xi)}\rightarrow 0 $ as $|\xi|\rightarrow 0$.
For the first part, I'm thinking to show that $|\hat{f}(\xi+h)-\hat{f}(\xi)|$ goes to zero when $h$ goes to zero. So, I have
$$ |\hat{f}(\xi+h)-\hat{f}(\xi)|=\left|\int_{\mathbb{R}}f(x)e^{-2\pi ix(\xi+h)}dx-\int_{\mathbb{R}}f(x)e^{-2\pi ix\xi}dx \right| $$
$$\leq\int_{\mathbb{R}} |f(x)|\left|e^{-2\pi ix\xi}\right|\left|e^{-2\pi ixh}-1\right|dx $$
If $h\rightarrow 0$, so $e^{-2\pi ixh}\rightarrow 1$, and this implies $\left|e^{-2\pi ixh}-1\right|\rightarrow 0$. So, the integral goes to zero and $\hat{f}$ is continuous. 
Is this proof correct?
For the second part, the hint is to prove the following statement:
$$\hat{f}(\xi)=\frac{1}{2}\int_{\mathbb{R}}\left[f(x)-f\left(x-\frac{1}{2\xi}\right)\right]e^{-2\pi i x \xi}dx $$
If that is true, when $|\xi|\rightarrow 0$, $f\left(1-\frac{1}{2\xi}\right)\rightarrow f(x)$, and it implies $\hat{f}(\xi)\rightarrow 0$. So, I just need to prove that statement.
All that I have is: 
$$\frac{1}{2}\int_{\mathbb{R}}\left[f(x)-f\left(x-\frac{1}{2\xi}\right)\right]e^{-2\pi i x \xi}dx=\frac{1}{2}\hat{f}(\xi)-\frac{1}{2}\int_{\mathbb{R}}f\left(1-\frac{1}{2\xi}\right)e^{-2\pi i x \xi}dx$$
So, prove the statement is equivalent to prove that 
$$\int_{\mathbb{R}}f\left(1-\frac{1}{2\xi}\right)e^{-2\pi i x \xi}dx=-\frac{1}{2}\hat{f}(\xi)$$
How can I prove that?
 A: If by moderate decrease you mean that $\int_{-\infty}^\infty |f(x)| \, dx < \infty$ then your proof of the first part is correct and the interchange of limit and integral is justified by the dominated convergence theorem.
For the second part, using the change of variables $x = u + \dfrac{1}{2\xi}$ we have, since $e^{-\pi i} = -1$,
$$\int_{-R}^R f\left(x- \frac{1}{2\xi}\right) e^{-2\pi i \xi x} \, dx = \int_{-R-\frac{1}{2\xi}}^{R- \frac{1}{2\xi}} f(u) e^{-2\pi i \xi u}e^{-\pi i} \, du  = -\int_{-R-\frac{1}{2\xi}}^{R- \frac{1}{2\xi}} f(u) e^{-2\pi i \xi u} \, du, $$
Thus,
$$\tag{*}\int_{-\infty}^\infty f\left(x- \frac{1}{2\xi}\right) e^{-2\pi i \xi x} \, dx = -\lim_{R\to \infty}\int_{-R-\frac{1}{2\xi}}^{R- \frac{1}{2\xi}} f(u) e^{-2\pi i \xi u} \, du = -\int_{-\infty}^{\infty} f(u) e^{-2\pi i \xi u} \, du  \\ - \hat{f}(\xi)$$
Subtracting (*) from $\displaystyle\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i \xi x} \, dx $, we get
$$2\hat{f}(\xi) = \int_{-\infty}^\infty f\left(x\right) e^{-2\pi i \xi x} \, dx - \int_{-\infty}^\infty f\left(x- \frac{1}{2\xi}\right) e^{-2\pi i \xi x} \, dx, $$
which implies
$$\hat{f}(\xi) = \frac{1}{2} \int_{-\infty}^\infty \left[f(x) - f\left(x- \frac{1}{2\xi}\right)\right] e^{-2\pi i \xi x} \, dx$$
