Fourier Transform on $L^1$ is not surjective.

Question

Let $f\in L^1(\mathbb{R})$ and suppose that $\hat f$ is an odd function.

Prove that there exists $C> 0$ s.t. for each $a>1$ we have $$ \left | \int_1 ^a \frac{\hat f(x)}{x} dx \right| \le C$$

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I have some observations:

1) If $\hat f$ is odd then $f$ is odd.

2) This hold for characteristic functions of intervals which are dense in $L^1$

Answer 1

$HINT$

Use the fact that for all $0<\epsilon<t< +\infty$ we have that $$|\int_{\epsilon}^t\frac{sinx}{x}dx| \leq 4$$

If you want you can prove it as an exercise.

Using the above fact prove that $$|\int_{\epsilon}^t\frac{\hat{f}(x)}{x}dx| \leq 4||f||_1$$

Then take an odd continuous function $g$ which is equal to $\frac{1}{\log{x}}$ for $x \geq 2$ to prove that there does not exist an $L^1$ function which has $g$ as its Fourier transform.