Let $(v_n)_n$ be an orthonormal base of $\mathcal H$ Hilbert space, $(\sigma_n)_n \subset \mathbb R$ and $x,y \in \mathcal H$.
Is $\sum_{n=1}^{\infty} \langle x, v_n \rangle \sigma_n^2v_n = y$ if and only if $x = \sum_{n=1}^{\infty} \langle y, v_n \rangle \sigma_n^{-2}v_n$?
Since $(v_n)_n$ is an orthonormal basis one can write any $x,y\in \mathcal{H}$ as a Fourier sum $$x=\sum_{n=1}^{\infty}\langle x,v_n\rangle v_n\\y=\sum_{n=1}^{\infty}\langle y,v_n\rangle v_n$$ Suppose that $(\sigma_n)_n\subset \mathbb{R}$ is such that $$y=\sum_{n=1}^{\infty}\sigma^2_n\langle x,v_n\rangle v_n$$ then $$\langle y,v_k\rangle=\langle \sum_{n=1}^{\infty}\sigma^2_n\langle x,v_n\rangle v_n,v_k\rangle=\sigma_k^2\langle x, v_k\rangle$$ Hence $$\sigma^{-2}_k\langle y,v_k\rangle=\langle x, v_k\rangle\Rightarrow x=\sum_{k=1}^{\infty}\langle x,v_k\rangle v_k=\sum_{k=1}^{\infty}\sigma^{-2}_k\langle y,v_k\rangle v_k$$ Analogue for the other direction. They are therefore equivalent. Note that $\sigma_n\neq 0$ for all $n$ is needed.