How to prove this Fourier question?

Question

How to prove this Fourier question? I hope for a procedure in detail.

Answer 1

You should assume something about $f$, otherwise why is $f_h$ even defined? Assuming $f$ is integrable on bounded intervals, we at least can find $a_n$ and $b_n$ by means of integrals involving $f$. When calculating the coefficients for $f_h$, introduce the variable $y=x+s$ $$\int_0^{2\pi} f_h(x)\cos nx\,dx = \frac{1}{2h}\int_0^{2\pi} \int_{-h}^h f(x+s)\cos nx\,ds\,dx \\ = \frac{1}{2h}\int_0^{2\pi} \int_{-h}^h f(y)\cos (ny-ns)\,ds\,dy$$ and use trigonometric identity $\cos (ny-ns)=\cos ny \cos ns + \sin ny\sin ns$. Now observe that integration over $s$ can be done holding $y$ constant: $$\int_{-h}^h \cos ns \,ds = \frac{2}{n}\sin nh,\qquad \int_{-h}^h \sin ns \,ds=0$$ Therefore, $$\int_0^{2\pi} f_h(x)\cos nx\,dx = \frac{\sin nh}{nh}\int_0^{2\pi} f(y)\cos ny\,dy$$

The computation for $\int_0^{2\pi} f_h(x)\sin nx\,dx$ is similar.

You did not specify in what sense the Fourier series for $f_h$ should converge to $f_h$, or if the proof of convergence is required at all. These lecture notes on pointwise and uniform convergence of Fourier series contain such results with detailed proofs.