# Stein's extension operator and wave front sets

Let $K\subset\mathbb{R}^d$ be a compact set with non-empty interior and Lipschitz boundary. In Section VI.3 of his book "Singular Integrals and Differentiability Properties of Functions", E. M. Stein constructs a linear operator $E_K$ continuously mapping the Sobolev space $W_{p,k}(K)$ into the Sobolev space $W_{p,k}(\mathbb{R}^d)$ for all $1\leq p\leq\infty$, $k=0,1,2,\ldots$, and such that $(E_K f)(x)=f(x)$ for all $x\in K$, $f\in W_{p,k}(K)$, nowadays called Stein's extension operator in the literature (actually, Stein builds $E_K$ with the above properties for $K$ closed with non-empty interior and locally Lipschitz boundary, but this will not be needed for the question). Consider now the linear operator $\tilde{E}_K:C^\infty(\mathbb{R}^n)\rightarrow C^\infty(\mathbb{R}^n)$ given by

$$\tilde{E}_K f=E_K(f|_K)$$

By Sobolev's embedding theorem (which does hold for $K$ as above), $\tilde{E}_K$ is a continuous linear map.

Question: Does the distribution kernel of $\tilde{E}_K$ have its wave front set contained in the conormal bundle to the diagonal of $\mathbb{R}^n\times\mathbb{R}^n$? In other words, does the extension of $\tilde{E}_K$ to $\mathscr{D}'(\mathbb{R}^n)$ by duality decrease wave front sets?

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The answer is NO already in one dimension. In fact, if $K$ is the interval $[-1, 1] \subset \mathbb R$ then the Schwartz kernel $K(x, y)$ of the operator $\tilde E_K: C^{\infty}(\mathbb R) \to C^{\infty}(\mathbb R)$ must be equal to $\delta(x-y)$ in the square $(-1, 1) \times (-1, 1)$, and its support must be contained in the union of that diagonal and the set $\{(x, y); |y| \le 1, |x| \ge 1 \}$. I claim that the wave front set of $K$ above the point $(1, 1)$ contains (at least) the entire angular region $-1 \le \xi/\eta \le 0$. To see this consider for $0 < a < 1$ the function $F(t) = \int _0^2 K(x, a(x-1) + t) dx$ in a neighborhood of $t=1$. For $t>1$ we have $F(t) = 0$, since the line $y = a(x-1) + t$ does not meet the support of $K$ then. If $t < 1$ this line intersects the diagonal, so the distribution $\delta(x-y)$ contributes by the amount $\int \delta((a-1)x) dx = 1/(1-a)$ to $F(t)$. In the region $\{(x, y); |y| \le 1, |x| \ge 1 \}$ the kernel $K(x, y)$ is an integrable function (in fact piecewize smooth). Hence $F(t) = 1/(1-a) + \mathcal O(|t-1|)$ as $t \to 1-0$, so $F(t)$ must have a jump at $t=1$. Therefore both conormals to the line $y = a(x-1) + 1$ at $(1,1)$ must belong to $WF(K)$. Since this is true for all $a \in (0, 1)$ and the wave front set is closed, this proves the claim.
Addendum: In the last step above I have used the fact that $F(t)$ must be smooth at $t_0$, if all conormals to the line $L(t_0): y = a(x-1) + t_0$ are absent in $WF(K)$; this follows directly from the definition of the wave front set using a partition of unity; see also Duistermath, Fourier Integral Operators, Proposition 1.3.4. In the same way we see that $(t_0, +1) \notin WF(F)$, if all conormals to $L(t_0)$ with the corresponding direction are absent in $WF(K)$, and the same is true for $(t_0, -1)$. The fact that $F(t)$ has a jump at $t=1$ implies that both cotangent vectors above $t=1$ belong to $WF(F)$. The support of $K$ meets the line $L(1)$ at just one point, $(1, 1)$. Hence both conormals to $L(1)$ at $(1,1)$ must belong to $WF(K)$.