Basic Properties Of Fourier Series. 2.

In this excercise we show how the symmetries of a function imply certain properties of its Fourier coefficients. Let $f$ be a 2$\pi$ Riemann integrable function defined on $R$.

(a) Show that the Fourier series of the function $f$ can be written as
$$f(\theta) \thicksim \hat{f}(0) + \sum_{n\ge1}[\hat{f}(n) + \hat{f}(-n)]\ cos n\theta + i[\hat{f}(n) – \hat{f}(-n)]\ sin n\theta. $$

(b)Prove that if $f$ is even, then $ \hat{f}(n)= \hat{f}(-n) $
, and we get a cosine series.

(c)Prove that if $f$ is odd, then $ \hat{f}(n)= -\hat{f}(-n) $, and we get a sine series.[This hint is helpful here Series Of Sines.

(d)Suppose that $f(\theta + \pi )$ = $f(\theta)$ for all $\theta \in R$. Show that $\hat{f}(n)$ = 0 for all odd n.(Its answer is here Effect of Symmetry of a Function on its Fourier Series)

(e)Show that $f$ is real-valued iff $\bar{\hat{f}}(n) =\hat{f}(-n)$ for all n.

For (e), we have two steps to prove 1- If $f$ is real valued, then $\bar{\hat{f}}(n) =\hat{f}(-n)$ for all n.Which is sooo easy by using the definition of Fourier Coeffient.

2- If $\bar{\hat{f}}(n) =\hat{f}(-n)$ for all n, Then $f$ is real valued.I reached this step,$$\int_{-\pi}^{\pi}\overline{f}(\theta)e^{in\theta} d\theta = \int_{-\pi}^{\pi}f(\theta)e^{in\theta} d\theta$$, Does this means that f is real valued ? If so what is the justification.

Solutions Collecting From Web of "Basic Properties Of Fourier Series. 2."

Let’s solve (a).

The Fourier series of the function $f$ is
$$
\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{in\theta}
$$

By definition, this is
$$
\sum_{n=0}^{\infty}\hat{f}(n)e^{in\theta} + \sum_{n=1}^{\infty}\hat{f}(-n)e^{i(-n)\theta}
$$

Taking out the $n=0$ term in the first series, this is
$$
\hat{f}(0)+\sum_{n=1}^{\infty}\hat{f}(n)e^{in\theta} + \sum_{n=1}^{\infty}\hat{f}(-n)e^{i(-n)\theta}\tag{1}
$$

Applying Euler’s formula we have
$$
e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\\
e^{i(-n)\theta}=\cos((-n)\theta)+i\sin((-n)\theta)
$$

Since $\cos$ is an even function and $\sin$ is an odd function, this is
$$
e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\\
e^{i(-n)\theta}=\cos(n\theta)-i\sin(n\theta)
$$
Substituting in $(1)$, we obtain
$$
\hat{f}(0)+\sum_{n=1}^{\infty}\hat{f}(n)[\cos(n\theta)+i\sin(n\theta)] + \sum_{n=1}^{\infty}\hat{f}(-n)[\cos(n\theta)-i\sin(n\theta)]
$$
But this equals
$$
\hat{f}(0)+\sum_{n=1}^{\infty}\left\{\hat{f}(n)[\cos(n\theta)+i\sin(n\theta)] + \hat{f}(-n)[\cos(n\theta)-i\sin(n\theta)]\right\}
$$
which is nothing more than
$$
\hat{f}(0)+\sum_{n=1}^{\infty}\left\{[\hat{f}(n) + \hat{f}(-n)]\cos(n\theta) + i[\hat{f}(n) – \hat{f}(-n)]\sin(n\theta)\right\}
$$