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We put, $\mathcal{S}(\mathbb R^{d})=$ The Schwartz space and $\mathcal{S'}(\mathbb R^{d})=$ The dual of $\mathcal{S}(\mathbb R^{d})$(The space of tempered distributions).

Suppose $\alpha > 1$ and $(\psi, \chi)$ be a couple of smooth functions valued in $[0,1],$ such that $\psi$ is supported in the shell $\{\xi \in \mathbb R^{d}:\alpha^{-1} \leq |\xi| \leq 2\alpha \},$ $\chi$ is supported in the ball $\{\xi \in \mathbb R^{d}: |\xi|\leq \alpha \}$ and $$\chi(\xi)+ \sum_{q\in \mathbb N} \psi (2^{-q}\xi)=1; \ \forall \xi \in \mathbb R^{d}.$$

For $f\in \mathcal{S'}(\mathbb R^{d}),$ one can define non-homogeneous dyadic blocks as follows. Let $\triangle_{q}f:=0, $ if $q\leq -2,$ $\triangle_{-1}f:=\chi(D)f= h\ast f$ with $h=\mathcal{F}^{-1}\chi,$ and $$\triangle_{q}f:=\psi(2^{-q}D)f= 2^{qd}\int_{\mathbb R^{d}}h(2^{q}y) f(x-y) dy \ \text{with} \ h=\mathcal{F}^{-1}\psi, (q\geq 0).$$

My Question is: Suppose that $\{f_{n}\}_{n\in \mathbb N}\subset \mathcal{S'}(\mathbb R^{d})$ converging to $f\in \mathcal{S'}(\mathbb R^{d}),$ that is, $$\lim_{n\to \infty} \left\langle f_{n}, \phi \right\rangle = \left\langle f, \phi \right\rangle;$$ for all $\phi \in \mathcal{S}(\mathbb R^{d}).$ Then, for each $q\in \mathbb Z,$ can we expect to show, $\{\triangle_{q}f_{n}\}_{n\in \mathbb N}$ converging to $\triangle_{q}f$ in $\mathcal{S'}(\mathbb R^{d})$ ? If yes, how ?

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    $\begingroup$ Isn't your question something like: If $(f_n)_n\to f$ in $\mathcal{S}'$, then $(k\ast f_n)_n\to k\ast f$ as long as $k$ is convolutable in $\mathcal{S}'$ ? $\endgroup$ – Vobo Aug 28 '14 at 18:45
  • $\begingroup$ Thanks; yes, but how does it help here ? (Actually, $\triangle_{q}f= 2^{qd}(k\ast f)$ ;where $k(y)=h(2^{q}y)$) $\endgroup$ – Inquisitive Aug 29 '14 at 12:53
  • $\begingroup$ As you see from the answer below, removing all irrelevant properties like "non-homogeneous dyadic blocks" makes it much easier to understand and prove the key facts of your problem. Reducing problems to a core question are the main steps when you want to apply mathematical methods in other disciplines. $\endgroup$ – Vobo Aug 30 '14 at 12:04
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There are basically two definitions of convolution of a distribution $F \in \mathcal{S}'$ with a Schwartz function $\psi$, which turn out to be equivalent (cf. Folland, Real Analysis, Proposition 9.10).

The first definition is to define $F \ast \psi$ as a pointwise-defined function(!),

$$ F \ast \psi (x) = \langle F, \tau_x \widetilde{\psi}\rangle, $$

where $\tau_x g(y) = g(y-x)$ and $\widetilde{g}(y) = g(-y)$. One can show that the function thus defined is smooth ($C^\infty$) and of polynomial growth in all derivatives, hence a tempered distribution.

The second definition is to define

$$ \langle F\ast\psi, \phi \rangle := \langle F, \phi \ast \widetilde{\psi} \rangle. $$

One then has to verify that the functional $F \ast \psi$ thus defined is indeed a tempered distribution.

For your question, the second definition (which is equivalent to the first, namely $\int (F\ast \psi)(x) \,dx = \langle F, \phi \ast \widetilde{\psi}\rangle$ holds) is more convenient, because if $F_n \to F$ in $\mathcal{S}'$, this implies

$$ \langle F_n \ast \psi, \phi \rangle = \langle F_n, \phi \ast \widetilde{\psi} \rangle \to \langle F, \phi \ast \widetilde{\psi}\rangle = \langle F \ast \psi, \phi \rangle $$

for all $\phi \in \mathcal{S}$, which means nothing but $F_n \ast \psi \to F \ast \psi$ in $\mathcal{S}'$.

Now apply this to $\psi = k$ and note that the (constant) scalar factor $2^{qd}$ is irrelevant for the convergence.

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