Finding coefficients for $\sum_{n=0}^\infty B_{nm}\sin(nx)\cos(my) = 1$

Question

I'm trying to find the $B_{nm}$ that satisfy: $$\sum_{m=0}^\infty \sum_{n=1}^\infty B_{nm}\sin(nx)\cos(my) = 1$$ for $0<x<\pi$ and $0<y<\pi$

Multiplying both sides by $\sin(n'x)\cos(m'y)$ and integrating, the LHS becomes:

$$ \\B_{nm}\int_0^\pi \sin(nx)\sin(n'x)dx\int_0^\pi \cos(my)\cos(m'y)dy =$$ $\frac{\pi^2}{4}B_{n'm'}\space$ for $\space m' \ne 0$

$\frac{\pi^2}{2}B_{n'0}\space$ for $\space m' = 0$

The RHS is: $$\int_0^\pi\sin(n'x)dx\int_0^\pi \cos(m'y)dy = 0$$ Therefore, $B_{nm} =0$

I can't think of another way to solve this problem. If I multiply through by $\sin(n'x)\sin(m'y)$ and integrate instead, I get:

$$ \\B_{nm}\int_0^\pi \sin(nx)\sin(n'x)dx\int_0^\pi \cos(my)\sin(m'y)dy =$$ $\sum_{n=1}^\infty\frac{\pi n((-1)^{n+m}-1)}{2(m^2-n^2)}B_{n'm'}\space$

which seems impossible to isolate $B_{nm}$

Answer 1

Hint: Have another look at the $n'=1$ term on the right-hand side. And then look at the other odd-$n'$ terms.

Answer 2

Partial answer:

If you restrict the $B_{mn}$ to be square summable the it is not possible to do what you want.

To see why, let $\phi(x) = \sum_{n_1 =0}^\infty \sum_{n_2 =1}^\infty B_{n_1 n_2} \sin (n_1 x_1) \cos (n_2 x_2)$ and suppose that the condition holds. (The issue is that $n_2$ starts at $1$.)

Then we have $\phi(x) = \operatorname{sgn} x_1$ ae. on $[-\pi,\pi]^2$ hence for $x_1 \neq 0$ and $m \ge 1$ we have $\int_{-\pi}^\pi \cos (mt) \phi((x_1,t)) dt = \int_{-\pi}^\pi \cos (mt) \operatorname{sgn} x_1 dt =0$ from which we get $\sum_{n_1 =0}^\infty B_{n_1 m} \sin (n_1 x_1) = 0$ for ae. $x_1$. It follows from this that $B_{n_1 m} = 0$ for all $n_1 \ge 0, m \ge 1$, which is a contradiction.

More directly, you could expand $\phi$ in terms of the basis formed from the orthogonal functions $x \mapsto b_1(n_1 x_1) b_2 (n_2 x_2)$ where $b_1,b_2 \in \{\cos,\sin \}$ and note that the coefficients corresponding to $n_2=0$ are not all zero.