Judging whether a function is not in the range of Fourier transformation (1) First, I have to show that if f is an odd function that is integrable on   the rea line, then there exists a positive number M such that for any a,A (where A is bigger) the following holds.

(2) And using this fact, I have to show that the given function g is continuous but there is no integrable function whose fourier transformation is equal to g.

I have no idea at all how to approach the two questions... Could anyone help me with these?
 A: If $f$ is odd and integrable, then
$$
           g(x) = \int_{0}^{x}f(y)dy
$$
is even and bounded with limits at $\pm\infty$. Therefore,
\begin{align}
         \int_{-R}^{R}e^{-isx}f(x)dx & = e^{-isx}g(x)|_{x=-R}^{R}+is\int_{-R}^{R}e^{-isx}g(x)dx \\
               & = -2i\sin(sR)g(R)+is\int_{-R}^{R}e^{-isx}g(x)dx. \\
        \frac{1}{2is}\int_{-R}^{R}e^{-isx}f(x)dx & = -\frac{\sin(sR)}{s}g(R)+\int_{0}^{R}\cos(sx)g(x)dx.
\end{align}
Integrate both sides over $[a,A]$ assuming $0\notin [a,A]$:
$$
       \int_{a}^{A}\frac{1}{2is}\int_{-R}^{R}e^{-isx}f(x)dxds \\
           = -\int_{a}^{A}\frac{\sin(sR)}{s}dsg(R)+\int_{0}^{R}\frac{\sin(Ax)-\sin(ax)}{x} g(x)dx \\
     = -\int_{a/R}^{A/R}\frac{\sin(u)}{u}du g(R)+\int_{0}^{R}\frac{\sin(Ax)-\sin(ax)}{x} g(x)dx
$$
Now suppose that $0 \notin [a,A]$. The integral on the left converges as $R\rightarrow\infty$ because the integrand converges uniformly to $\sqrt{2\pi}\hat{f}(s)/2is$. The evaluation terms on the far right converge to $0$ as $R\rightarrow\infty$ because $\int_{0}^{x}\frac{\sin(u)}{u}du$ is uniformly bounded in $x$ and converges to $0$ as $x\rightarrow 0$, and $g(x)$ is uniformly bounded. Therefore, the remaining integral on the right must also converge as $R\rightarrow\infty$. Hence, if $0\notin[a,A]$,
$$
           \sqrt{2\pi}\int_{a}^{A}\frac{1}{2is}\hat{f}(s)ds
       = \int_{0}^{\infty}\frac{\sin(Ax)-\sin(ax)}{x}g(x)dx.
$$
The right side is guaranteed to exist as an improper integral. Integrating the right side by parts gives
$$
    \left.\int_{0}^{x}\frac{\sin(Ay)-\sin(ay)}{y}dyg(x)\right|_{x=0}^{\infty}
   - \int_{0}^{\infty}\int_{0}^{x}\frac{\sin(Ay)-\sin(ay)}{y}dy f(x)dx \\
    = -\int_{0}^{\infty}\int_{0}^{x}\frac{\sin(Ay)-\sin(ay)}{y}dy f(x)dx
$$
which is uniformly bounded by
$$
               2\int_{0}^{\pi/2}\frac{\sin(y)}{y}\int_{0}^{\infty}|f(x)|dx
              = \int_{0}^{\pi/2}\frac{\sin(y)}{y}\int_{-\infty}^{\infty}|f(x)|dx.
$$
So it appears to me that
$$
     \left|\int_{a}^{A}\frac{\hat{f}(s)}{s}ds\right|\le\sqrt{\frac{2}{\pi}}\int_{0}^{\pi/2}\frac{\sin(x)}{x}dx\int_{-\infty}^{\infty}|f(x)|dx.
$$
So I think the best constant $M$ is
$$
               M = \sqrt{\frac{2}{\pi}}\int_{0}^{\pi/2}\frac{\sin(x)}{x}dx\int_{-\infty}^{\infty}|f(x)|dx
$$
I've tried to be careful about the details because I'm a little surprised that such a result can be true. Please check and ask any questions that you have. I don't think the interval $[a,A]$ can include $0$; even though the integral remains uniformly bounded for $0 < a < A$ as $a \downarrow 0$, I doubt that it can converge in general as $a\downarrow 0$.
For the $g$ as stated, suppose that $g(s)=\hat{f}(s)$ for some absolutely integrable $f$. Then
$$
                    g(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ist}f(t)dt
$$
automatically gives the following by a change of variable
\begin{align}
        g(s) & = \frac{g(s)-g(-s)}{2} \\
             & =\frac{1}{2}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(e^{-ist}-e^{ist})f(t)dt \\
             & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
                   e^{-ist}\frac{f(t)-f(-t)}{2}dt.
\end{align}
Therefore, because $g$ is odd, it can be assumed that $g$ is the Fourier transform of an odd absolutely integrable function. So the above result applies.
But notice that the following is not uniformly bounded in $A$:
$$
               \int_{1}^{A}\frac{g(\alpha)}{\alpha}d\alpha=\int_{1}^{A}\frac{\ln(\alpha)}{\alpha}d\alpha 
         = \int_{0}^{\ln(A)}udu
$$
