Complex form of Fourier's Theorem
Any periodic function can be decomposed into a linear combination of complex exponentials.
Proof
Consider a complex exponential with period $T_0$: $$e^{i\frac {2π}{T_0}t}$$ Hence, we have: $$\int^{T_0}_0 e^{i\frac {2π}{T_0}t} dt = 0 ……(1)$$ Let us claim that for a periodic function $x(t)$ we can write: $$x(t) ≈ \sum^{N}_{k=-N} C_ke^{i\frac {2πk}{T_0}t},$$ where $2N+1$ is the number of frequency components used. As $N→∞$, we have: $$x(t) = \sum^{∞}_{k=-∞} C_ke^{i\frac {2πk}{T_0}t} ……(2)$$ Consider: $$v_k(t):=e^{i\frac {2πk}{T_0}t}……(3)$$ Then we have: $$v_k(t+T_0)=v_k(t)$$ Hence, $v_k(t)$ is a periodic function. Now, $$\int^{T_0}_0 v_k(t)v^*_l(t) dt$$ $$=\int^{T_0}_0 e^{i\frac {2π(k-l)}{T_0}t} dt$$ $$=0 (k≠l)$$ $$T_0 (k=l)……(4)$$ Thus, $v_k(t)$ is orthogonal. Moreover, if we assume equation (2) to be valid , then we can multiply both its sides by $v_l^*(t)$ and integrate over $T_0$ we get: $$C_k=\frac {1}{T_0} \int^{T_0}_0 x(t)v_k^*(t) dt ……(5)$$ Putting this value in equation (2) we get back $x(t)$. Hence, any periodic function can be decomposed into a linear combination of complex exponentials.
Real form of Fourier's Theorem
An arbitrary periodic function $F(t)$ with period $T$ can be decomposed into a linear combination of the functions $f_n(t)$ and $g_n(t)$ where, $$f_n(t)=\sin \frac {2πnt}{T}$$ $$g_n(t)=\cos \frac {2πnt}{T}$$ Mathematically, $$F(t)=b_0 + b_1g_1(t) +b_2g_2(t) + …… + a_1f_1(t) +a_2f_2(t) + ……,$$ where $n$ is a non-negative integer and all of $a_i$,$b_i$ are real.
Problem
Is there a similar proof for the real form of Fourier's Theorem as for the complex form? I couldn't get one on the internet or in any book. Does this mean that it can be derived from the complex form? If yes then how? Any help would be appreciated.
The functions $$ 1,\sin(2\pi x/T),\cos(2\pi x/T),\sin(4\pi x/T),\cos(4\pi x/T),\cdots $$ are mutually orthogonal. That is, if you choose any two different functions $f,g$ from the above, then $$ \int_{-T}^{T} f(x)g(x)dx = 0. $$ And, $$ \int_{-T}^{T}1^2dx=2T\\ \int_{-T}^{T}\sin^2(2\pi x/T)dx=T\\ \int_{-T}^{T}\cos^2(2\pi x/T)dx=T. $$ That's enough to carry about the same argument that you used before. It should be noted that your argument assumes convergence of the series, and does not prove it.