Let $(V, \langle\cdot,\cdot\rangle)$ be an infinite dimensional inner product space.
Gram-Schmidt says that for all $n$ we can find $n$ vectors $v_1,...,v_n$ so that $\langle v_i,v_j\rangle=0$ for all $i\neq j$. I am curious - can we change the $0$? That is:
Can we find $n$ distinct vectors so that $\langle v_i,v_j\rangle=4$ for all $i\neq j$, for example?
It is tricky. Gram-Schmidt just uses bilinearity which is insufficient here - as the zero bilinear form shows that the answer to the question above is no if we just require bilinearity.
Let's suppose that you find the regular orthonormal basis $u_1, ..., u_n$, with $\langle u_i,u_j\rangle=\delta_{i,j}$. Then let's construct $$U=\sum_{i=1}^nu_i$$ Notice that $\langle U,U\rangle=n$. Now add $CU$, $C\in\mathbb R$, to the orthonormal basis vectors $$v_i=u_i+CU$$ Then you have $$\langle v_i, v_j\rangle=\langle u_i+CU, u_j+CU\rangle\\=\langle u_i, u_j\rangle+\langle u_i, CU\rangle+\langle CU, u_j\rangle+\langle CU, CU\rangle\\=\delta_{i,j}+2C+C^2n$$ Now just choose $2C+C^2n=4$ and you get your desired vectors.