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?

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

1 Answer 1


(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.


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