The problem I am considering is:
For $f$ a $2\pi$-periodic and Riemann integrable function, show that $f$ is real valued iff $\overline{ \hat{f}(n) } = \hat{f}(-n)$.
Here $\hat{f}(n)$ represents the $n$th fourier coefficient.
It is easy to see that $f$ real implies the equality.
Going the other direction, all I am able to deduce is that $\widehat{\bar{f} - f}(n) = 0$ $\forall n \in \mathbb{Z}$, which suggests that $f$ is real whenever it is continuous. (Since the Fejer kernel is a good kernel, the Fourier series is Cesaro summable to every point of continuity of $f$.)
I know that I can approximate in $L_1$ norm an integrable function on the disc. However, all this would give (provided that I could translate the equality in terms of the approximating sequence) is that $f$ is real a.e. This also seems to be a dead end.
Can someone suggest an alternative approach?
(I am reading Stein and Shakarachi - Fourier Analysis.)
In the next chapter they discuss the convergence of the partial sums in $L^2$, but I do not think I am allowed to use that. (Again, that would only provide real valued a.e.)
Strictly speaking, this isn't true as stated. Let $f$ be $0$ everywhere, except $f(2\pi k)=i$. Then $f$ is Riemann integrable, but the Riemann integral won't pick up the non-real value at the single point, so $\hat f(n)=0$ for all $n$. Real-valued almost everywhere is the best you can hope for.