Given two sequences of biorthogonal functions: $$\left \{ f_{n} \right \}, \left \{ g_{n} \right \}\\\left \langle f_{n},g_{m} \right \rangle=\delta_{nm}\\\left \langle f_{n},f_{m} \right \rangle\neq \delta_{nm} \\ \left \langle g_{n},g_{m} \right \rangle\neq \delta_{nm}$$ is it possible to find a linear combination of the two sequences such that they are converted into complete orthogonal ones? i.e, could there be two sequences of constants $\left \{ a_{n} \right \}$, $\left \{ b_{n} \right \}$ such that : $$\left \langle a_{n}f_{n}+b_{n}g_{n},a_{m}f_{m}+b_{m}g_{m} \right \rangle=\delta_{nm}$$ In other words, do there exist two sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ such that :
$$a_{n}a_{m}C_{nm}+b_{n}b_{m}D_{nm}+\left(a_{n}b_{m}+a_{m}b_{n} \right )\delta_{nm}=\delta_{nm}\\ C_{nm}=\left \langle f_{n},f_{m} \right \rangle \\ D_{nm}=\left \langle g_{n},g_{m} \right \rangle $$
We must have for $m\neq n$ that $$a_ma_nC_{mn}+b_mb_nD_{mn}=0\qquad\qquad(1)$$and for $m=n$$$a_n^2C_{nn}+b_n^2D_{nn}=1-2a_nb_n\qquad\qquad(2)$$We show that this is possible if and only if $$D_{mn}=-f_mf_nC_{mn}$$for some $f_n$ when $m\neq n$.
Clearly, $(1)$ leads to$$\dfrac{a_m}{b_m}\dfrac{a_n}{b_n}C_{mn}+D_{mn}=0$$which leads to $$D_{mn}=-\dfrac{a_m}{b_m}\dfrac{a_n}{b_n}C_{mn}$$thereby taking $f_n=\dfrac{a_n}{b_n}$ we have proven one side of the theorem. The proof of converse can also be followed easily. Let's assume $$a_n=b_n\sqrt{-\dfrac{D_{nn}}{C_{nn}}}$$by substituting in (2) we obtain$$2b_n^2\sqrt{-\dfrac{D_{nn}}{C_{nn}}}=1$$or $$b_n=\dfrac{1}{\sqrt 2}\left(-\dfrac{C_{nn}}{D_{nn}}\right)^{\dfrac{1}{4}}\\a_n=\dfrac{1}{\sqrt 2}\left(-\dfrac{D_{nn}}{C_{nn}}\right)^{\dfrac{1}{4}}$$