# Is $\mathcal{E}' \ni u \mapsto \hat{u} \in C^\infty$ sequentially continuous?

Let $\Omega \subseteq \mathbb{R}^n$ be open. For any compactly supported distribution $u \in \mathcal{E}'(\Omega)$, the distributional Fourier transform $\hat{u}$ is in fact a $C^\infty$ function on $\mathbb{R}^n$ with formula given by $$\hat{u}(\xi) = \langle u(x), \chi(x) e^{i x \cdot \xi} \rangle,$$

where $\chi$ is any element of $C_0^\infty(\mathbb{R}^n)$ such that $\chi \equiv 1$ on the support of $u$.

Suppose we have a sequence $\{u_j\}_{j = 1}^\infty \subseteq \mathcal{E}'(\Omega)$ along with $u \in \mathcal{E}'(\Omega)$ such that the supports of the $u_j$ and of $u$ are contained in a fixed compact set $K \subseteq \Omega$. Moreover, suppose that $$\langle u_j, \varphi \rangle \to \langle u, \varphi \rangle \qquad \text{as } j \to \infty$$ for all $\varphi$ in the Schwartz Class $\mathcal{S}(\mathbb{R}^n)$.

I would like to know, given any multi-index $\alpha$ and any compact $K' \subseteq \mathbb{R}^n$, must it be the case that the $D^\alpha_\xi \hat{u}_j$ converges uniformly to $D^\alpha_\xi \hat{u}_j$ on $K'$?

Note: it can be shown that $D_\xi^\alpha \hat{u}(\xi) = \langle u, (ix)^\alpha \chi(x) e^{ix\cdot \xi} \rangle$ for all multi-indices $\alpha$.

Edit: Here is my incomplete attempt to show that the statement is true. I guess that they best way to go is to try and use the uniform boundedness principle, but I may be wrong.

Every element of $\mathcal{E}'(\Omega)$ is a continuous linear map from the Fréchet space $C^\infty(\Omega)$ to $\mathbb{C}$. Note that the family of seminorms $\rho_m$, $m \in \mathbb{N}$, that defines the Fréchet space topology on $C^\infty(\Omega)$ is

$$\rho_m(\varphi) = \sup \{|\partial^\beta \varphi(x)| : x \in K_m, \text{ } \beta \le m \}, \qquad \varphi \in C^\infty(\Omega),$$ where $\{K_m\}_{m= 1}^\infty$ is a compact exhaustion of $\Omega$.

The convergence criterion supposed above guarantees that the continuous linear operators $T_j \equiv u_j-u : C^\infty(\Omega) \to \mathbb{C}$ have the property

$$\sup_j |T_j \varphi| < \infty, \qquad \text{for all \varphi \in C^\infty(\Omega).}$$

Therefore, the uniform boundedness principle for Fréchet spaces implies that the family $T_j$ is equicontinuous. This means that for every $\varepsilon > 0$ there exists $\delta_\varepsilon > 0$ and an $m_\varepsilon \in \mathbb{N}$ so that

$$\rho_{m_\varepsilon}(\varphi) < \delta_\varepsilon \implies |T_j(\varphi)| < \varepsilon, \qquad \text{for all j.}$$

So, for given $\varepsilon > 0$, we want to show that there exists $J \in \mathbb{N}$ so that

$$|D^\alpha_\xi (\hat{u}_j - u)(\xi) | = |T_j((ix)^\alpha \chi(x) e^{i x \cdot \xi})| < \varepsilon \qquad \text{all } \xi \in K' \text{ and all } j \ge J.$$

At this stage, my intuition is that the convergence in $\mathcal{E}'(\Omega)$ which I suppose is not strong enough to guarantee the convergence in $C^\infty(\mathbb{R}^n)$ I am seeking, because I do not have an effective way to make $(ix)^\alpha \chi(x) e^{i x \cdot \xi}$ small (uniformly in $\xi \in K'$) in the $\rho_\varepsilon$ seminorm.

Although I have yet to come up a with a counterexample.

Hints or solutions are greatly appreciated!

## 1 Answer

The Fourier transform $\mathcal E'\to C^\infty$ is continuous if $\mathcal E'$ is endowed with the strong topology $\beta(\mathcal E',\mathcal E)$ which is an inductive limit of Banach spaces with compact inclusion (a so-called LS- or Silva-space). This follows e.g. from the closed graph theorem. If $\langle u_j,\varphi\rangle$ converge for all $\varphi \in \mathcal S$ and the supports are fixed then these sequences converge for all $\varphi\in \mathcal E$ (by multiplying with $\chi$) which means that $u_j$ is $\sigma(\mathcal E',\mathcal E)$ bounded and hence $\beta(\mathcal E',\mathcal E)$-bounded. But bounded sets in $\mathcal E'$ are relatively compact so that the weak and strong topology coincide. Hence, the sequence $u_j$ converges in $\mathcal E'$ and by continuity $\hat u_j$ converges in $C^\infty$, i.e., all derivatives converge uniformly on compact sets.

• Thank you for the answer. I am not so familiar with the topologies you mention. Do you know of a reference that discusses distribution theory from this topological perspective? Also, I have added a slightly more concrete solution that attempts the uniform boundedness principle, but have gotten stuck. – JZS Mar 11 '18 at 20:47