Suppose $f \colon I \to \mathbb{R}$ is a function in, say, $L^\infty$, and $I \subset \mathbb{R}$ is a bounded interval. We may assume further regularity on $f$, such as Lipschitz continuity or strict positivity, in case it matters.

Consider all polynomials of $f$, such as $3f^4 - f^2 +2$. These are in $L^2(I)$, since $f$ is (essentially) bounded and $I$ is bounded. Let $P_f$ be the space of all such polynomials.

What can one say about the closure of $P_f$ in $L^2(I)$, that is $\overline{P_f}^{L^2(I)}$?

Extreme examples

1. If $f$ is constant (on a given subset of $I$), then so are all of its polynomials.
2. If $f(x) \equiv x$, then $P_f$ is the set of all polynomials, which is dense in $L^2(I)$.

Maybe the level sets of $f$ are the only obstruction for the closure to be all of $L^2$? I would be pleasantly surprised if this was the case.

Suppose $f \colon I \to \mathbb{R}$ is a function in, say, $L^\infty$, and $I \subset \mathbb{R}$ is a bounded interval. We may assume further regularity on $f$, such as Lipschitz continuity or strict positivity, in case it matters.

Consider all polynomials of $f$, such as $3f^4 - f^2 +2$. These are in $L^2(I)$, since $f$ is (essentially) bounded and $I$ is bounded. Let $P_f$ be the space of all such polynomials.

What can one say about the closure of $P_f$ in $L^2(I)$, that is $\overline{P_f}^{L^2(I)}$?

Examples

1. If $f$ is constant (on a given subset of $I$), then so are all of its polynomials.
2. If $f(x) \equiv x$, then $P_f$ is the set of all polynomials, which is dense in $L^2(I)$.
3. If $f$ has an inverse that is a polynomial, then it can be reduced to the previous case. Maybe this can be done also with functions whose inverse can approximated by polynomials in a suitable strong sense.

Maybe the level sets of $f$ are the only obstruction for the closure to be all of $L^2$? I would be pleasantly surprised if this was the case.

Suppose $f \colon I \to \mathbb{R}$ is a function in, say, $L^\infty$, and $I \subset \mathbb{R}$ is a bounded interval. We may assume further regularity on $f$, such as Lipschitz continuity or strict positivity, in case it matters.

Consider all polynomials of $f$, such as $3f^4 - f^2 +2$. These are in $L^2(I)$, since $f$ is (essentially) bounded and $I$ is bounded. Let $P_f$ be the space of all such polynomials.

What can one say about the closure of $P_f$ in $L^2(I)$, that is $\overline{P_f}^{L^2(I)}$?

Extreme examples

1. If $f$ is constant (on a given subset of $I$), then so are all of its polynomials.
2. If $f(x) \equiv x$, then $P_f$ is the set of all polynomials, which is dense in $L^2(I)$.

Suppose $f \colon I \to \mathbb{R}$ is a function in, say, $L^\infty$, and $I \subset \mathbb{R}$ is a bounded interval. We may assume further regularity on $f$, such as Lipschitz continuity or strict positivity, in case it matters.

Consider all polynomials of $f$, such as $3f^4 - f^2 +2$. These are in $L^2(I)$, since $f$ is (essentially) bounded and $I$ is bounded. Let $P_f$ be the space of all such polynomials.

What can one say about the closure of $P_f$ in $L^2(I)$, that is $\overline{P_f}^{L^2(I)}$?

Extreme examples

1. If $f$ is constant (on a given subset of $I$), then so are all of its polynomials.
2. If $f(x) \equiv x$, then $P_f$ is the set of all polynomials, which is dense in $L^2(I)$.

Maybe the level sets of $f$ are the only obstruction for the closure to be all of $L^2$? I would be pleasantly surprised if this was the case.