If I know $\langle g(f), a_i \rangle$ for all $i$ where $a_i$ is a basis, do I know the coefficients of $f$?

Question

Here I work on a bounded domain. Let $A=\text{span}(a_1, ..., a_n)$ where $\{a_i\}_{n=1}^\infty$ is a basis (not orthonormal) of $L^p \cap L^2$. Suppose $f \in A$.

Let $g:\mathbb{R} \to \mathbb{R}$ be a continuous invertible (nonlinear!) function that passes through the origin and has the property that $g:L^p \to L^{p'}$.

If I know $b_i := \langle g(f), a_i \rangle = \int g(f)a_i$ for each $i$, does that determine what $f$ is? I.e. can I write the coefficients of $f$ in terms of $b_i$? Note that this integral is well-defined.

Here $p$ and $p'$ are conjugate indices.

This is related to my last thread.

Answer 1

does that determine what $f$ is?

Not in general. It determines what $g\circ f$ is, because a continuous linear functional is determined by its values on a basis of the space. But the function $g$ could attain some values more than once (e.g., be constant on some interval), which would make it impossible to recover $f$ from $g\circ f$.

can I write the coefficients of $f$ in terms of $b_i$

No. Any nonlinear transformation of a function hopelessly messes up its coefficients in a basis. (An exception: Fourier coefficients behave reasonably when one squares a function, but even then it's still a convolution of a doubly infinite sequence with itself; not something one can explicitly compute in practice.)