(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?
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 $A$: $$ \int_{1}^{A}\frac{g(\alpha)}{\alpha}d\alpha=\int_{1}^{A}\frac{\ln(\alpha)}{\alpha}d\alpha = \int_{0}^{\ln(A)}udu $$