In Lemma 1.4.3 of Anton Deitmar's A First Course in Harmonic Analysis, he proves that the Fourier series of a periodic function $f:\mathbb R\to\mathbb R$ such that $\left.f\right|_{\left[0,1\right]}$ is a Riemann step function converges to $f$ in $L^2\left(\left[0,1\right]\right)$.
He makes extensive use of the equality
$$\left\|f\right\|_2^2=\sum_{n=-\infty}^\infty\left|c_n\right|^2,\tag{$\star$}$$
where the $c_n$s are the Fourier coefficients of $f$.
He first proves it for the case where $f\left(x\right)=\mathbf1_{\left[0,a\right]}\left(x\right)$, where $a\in\left[0,1\right]$, and then extends the proof to the case where $f\left(x\right)=\mathbf1_I\left(x\right)$, where $I$ is an open, closed, or half-open subinterval of $\left[0,1\right]$. I understand these proofs.
However, at the end he says that the proof extends, "by linearity", to the case where
$$f\left(x\right)=\sum_{n=1}^m\alpha_n\mathbf1_{I_n}\left(x\right),$$
where the $\alpha_n$s are constants. How is this so? It does not look like I can use "linearity" with $\left(\star\right)$.
You need to show that if $(*)$ holds for $f$ and $g$, then it holds for $\alpha f$ (obvious) and $f+g$ (not obvious). Let's stick to real-valued functions $f$ and $g$. Let $f$ have Fourier coefficients $c_n$ and $g$ have Fourier coefficients $d_n$.
We need to show that $\|f+g\|^2=\sum_n (c_n+d_n)^2$. From Bessel's inequality, $\|f+g\|^2\ge\sum_n (c_n+d_n)^2$ and $\|f-g\|^2\ge\sum_n (c_n-d_n)^2$. But $$\|f+g\|^2+\|f-g\|^2=2\|f\|^2+2\|g\|^2= 2\sum_n(c_n^2+d_n^2)=\sum_n(c_n+d_n)^2 +\sum_n(c_n-d_n)^2.$$ Using Bessel's inequality for $f-g$ gives $$\|f+g\|^2\le\sum_n(c_n+d_n)^2.$$ Another application of Bessel gives $$\|f+g\|^2=\sum_n(c_n+d_n)^2$$ so $(*)$ holds for $f+g$.
For complex valued functions use $\|f+g\|^2+\|f+ig\|^2+\|f-g\|^2+\|f-ig\|^2 =4\|f\|^2+4\|g\|^2$.