# 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?

## 1 Answer

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 $WF(K)$.

• I've repeated the steps of your argument and I think I've got it now. The point of your line integration along the second argument was just to display the additional conormal singularities of the kernel at boundary points, correct? It did take me some time to visualize this... Thank you very much for the patience and help! – Pedro Lauridsen Ribeiro Sep 6 '11 at 23:59