Let $T$ be a linear operator defined on the space of the algebraic polinomials in $[0,1]$ (polinomials with rational coefficients) such that for each $k \in \mathbb{N}, T[x^k]=0$. Is it possibile to extend the operator to the space of continuous functions in $[0,1]$? In that case, how would it value $T[sin(x)]$?
I was thinking about using Hahn-Banach theorem. Actually, since the subspace is dense in $C([0,1])$ it is possible to have a unique extension (I read this through my notes…even if I don't have any proofs of it; if someone can send me a link about it, this would be really appreciated…is this just extending the operator to the closure of the domain where is defined?).
After that, one corollary of Hahn-Banach theorem says:
if $Y \subset X$ is a closed subspace and $x \notin Y$ then it exists $\Lambda \in X^*$ such that $\Lambda x=1$ but $\Lambda y=0$ for all $y \in Y$ (in this case I know the proof).
This fact let me think that $T[sin(x)]=1$…am I right?
Best Answer
Let $P$ be the vector space of polynomials; since each polynomial is a finite linear combination of monomials and $T(x^k) = 0$ for all $k$, $T : P \to P$ is the zero operator. In particular, $T$ is continuous with respect to the sup norm over $[0,1]$ (constant functions are always continuous).
Now recall that $P$ is dense in $C([0,1])$. Clearly the zero operator is a continuous extension of $T$. Suppose $T'$ is another continuous extension of $T$. It's a standard fact from topology that if two continuous functions taking values in a Hausdorff space agree on a dense set, they are the same function. (In this case, note that $T-T'$ is continuous, therefore $(T-T')^{-1}(\{0\})$ is closed...) So this $T$ has a unique continuous extension to all of $C([0,1])$: namely, the zero operator.
More generally, if $X$ is a normed space, $Y$ is a Banach space, $E$ is a dense subspace of $X$, and $T : E \to Y$ is continuous, then there exists a continuous extension $T' : X \to Y$. This is often called the BLT theorem (for "bounded linear transformation"). By the argument we gave above, once the extension exists, it is necessarily unique.
Hahn-Banach covers the complementary case where $E$ is not dense. In this case, in general you are only guaranteed a continuous extension if $Y$ is finite dimensional, and the extension is never unique.