# Is $\mathscr{S}_h'$ a complementary subspace for $\mathscr{S}'/\mathscr{P}$, the space of tempered distributions modulo polynomials?

Recall that in many Fourier Analysis texts, given a function $\Psi$ such that $\hat{\Psi}\in\mathcal{D}(\mathbb R^d)$, $\hat\Psi\ge0$ is supported in an annulus, and $\sum_{j\in\mathbb Z}\hat\Psi(2^j\xi)=1$ for all $\xi\ne0$, we can define the Littlewood-Paley operator $$\Delta_j(f) := \left(\hat\Psi(2^{-j}\xi)\hat{f}(\xi)\right)^\vee$$ for $f\in\mathscr{S}'(\mathbb R^d)$. Furthermore, if we set $$\hat\Phi(\xi)=1-\sum_{j\ge0}\Psi(2^{-j}\xi)$$ then we can define $$\dot S_j(f):=\left(\hat\Phi(2^{-j}\xi)\hat f(\xi)\right)^\vee$$ and using this, we define $\mathscr{S}'_h(\mathbb R^d)$ to be the space of tempered distributions $u$ such that $\lim\limits_{j\to-\infty}\dot S_j(u)=0$ in $\mathscr{S}'(\mathbb R^d)$.

Now, clearly the space $\mathscr{P}$ of polynomials in $\mathscr{S}'$ intersects trivially with $\mathscr{S}_h'$, but is $\mathscr{S}_h'$ in fact a complementary subspace for $\mathscr P$? It would suffice to show that the limit $T(u):=\lim\limits_{j\to-\infty}\dot S_j(u)$ exists for any $u\in\mathscr{S}'$, since $T(u)\in\mathscr P$ and $T^2=T$ wherever $T$ is defined, which would make $T$ into a projection. However, I have only managed to show that $T(u)$ is defined when $\operatorname{ord}u=0$, i.e. when $u$ is locally a measure. But is $\mathscr{S}'_h$ even a complementary subspace, and if so, how do we show that?

• $pv(1/x)$ doesn't have such a decomposition Jan 15, 2020 at 5:06

(This is rather late, but I'll answer the question anyways.) No, it is not true that $$\mathscr S_h'$$ and $$\mathscr P$$ are (topological) complements in $$\mathscr S'$$ (in fact, they are not even linear complements). Suppose it were the case that every tempered distribution $$f$$ admitted a (unique) decomposition of the form $$f= g+P$$ for some $$g \in \mathscr S_h'$$ and some $$P \in \mathscr P$$. Fix a Schwartz function $$\varphi$$ with $$\varphi(0) \neq 0$$. Then, for any $$f \in \mathscr S'$$, $$\langle f, \dot S_j \varphi \rangle = \langle \dot S_j g , \varphi \rangle + \langle \dot S_j P, \varphi \rangle \to 0 + \langle P, \varphi \rangle$$ as $$j \to - \infty$$. This implies that the sequence $$( \dot S_j \varphi)_{j \leq 0}$$ is Cauchy for $$\sigma(\mathscr S, \mathscr S')$$. Now, $$\mathscr S$$ is a semi-Montel space (much more is true, but this is enough for our purposes), so $$\sigma(\mathscr S, \mathscr S')$$ is quasi-complete (hence sequentially complete), and $$\sigma(\mathscr S, \mathscr S')$$ convergent sequences also converge in the usual Fréchet topology on $$\mathscr S$$. As the Fourier transform is a linear homeomorphism on $$\mathscr S$$ (with its Fréchet topology), we see that the sequence $$(\hat \Phi(2^{-j} \cdot) \hat \varphi)_{j \leq 0}$$ converges in $$\mathscr S$$ as $$j \to - \infty$$. But, $$\partial_1 ( \hat \Phi(2^{-j} \cdot) \hat \varphi ) = 2^{-j} (\partial_1 \hat \Phi)(2^{-j} \cdot) \hat \varphi + \hat \Phi(2^{-j} \cdot) \partial_1 \hat \varphi.$$ The second term remains bounded in $$L^{\infty}$$ as $$j \to - \infty$$. Take some $$x_0$$ such that $$\partial_1 \hat \Phi(x_0) \neq 0$$. Then, $$|2^{-j} \partial_1 \hat \Phi(x_0) \hat \varphi(2^j x_0) | \to \infty$$ as $$j \to - \infty$$, resulting in a contradiction.