Well-defined distribution and its singular support

Question

Let $f$ be a smooth function on $X$, an open subset of $\mathbb{R}^n$, with $Im(f) \geq 0$. Let us fix an $\epsilon > 0$. Let $T_{\epsilon} := \frac{1}{f(x)+i\epsilon} $ in $D’(X)$.

Now if we assume $df(x) \neq 0$ when $f(x) = 0$, I wanna show that the limit of $T_{\epsilon}$ exists in $D’(X)$, that is $T :=\frac{1}{f(x)+i0}$ and then find its singular support and wavefront set.

For that, I showed that $$\frac{1}{f(x)+i\epsilon} = \int_{0}^{\infty} e^{i(f(x)+i\epsilon)t}dt ,$$ and I want to use this expression to show that $T$ is well-defined with the dominated convergence formula. But I think I am missing something because i don’t use the assumption $df(x) \neq 0$ when $f(x) = 0$ ; I don’t know how to put everything together.

Any ideas ? Thank you.

Answer 1

$\bf\text{Claim:}$ The wave-front-set of $\displaystyle T(x)=\frac{1}{f(x)+i0}$ is $$\bigl(Z_+\times (0,+\infty)\bigr)\cup \bigl(Z_-\times (-\infty,0)\bigr), \tag{1}$$ where $ Z_\pm=\{x\in X, f(x)=0, \pm \Re f'(x)>0\}. $

$[\mathbf 1]. $$\bf\text{To start the proof}$ of that claim, let us first show that $T$ is well-defined. Nothing is happening away from the zeros of $f$, so let us consider a point $x_0$ such that $f(x_0)=0$. With $f_1=\Re f, f_2=\Im f$, we have since $f_2$ is non-negative $$ f_j(x_0)=0, f'_1(x_0)\not=0, f'_2(x_0)=0. $$ We note in particular that when $x_0$ is a zero of $f$ we have $f'_1(x_0)\not=0$. As a result, assuming that $x_0=0$ we get with $a=f'_1(0)\not=0,$ $$ f(x)=a x+x^2 g(x), \quad\text{$g$ smooth complex-valued, $\Im g\ge 0$.} $$ Let us assume that $a>0$ and by a linear change of variable that $a=1$. Using the same notations, we have near $0$, with $g_1=\Re g, g_2=\Im g\ge 0$ $$ f(x)+i\epsilon=x(1+xg_1(x))+i(\epsilon+x^2g_2(x)). $$ We may change the variable near 0 and take $y=x(1+xg_1(x))$ as a new variable so that $$ F(y)+i\epsilon=f(x(y))+i\epsilon=y+i\bigl(\epsilon+y^2 g_3(y)\bigr), \quad g_3\ge 0. $$ Since the imaginary part of $F+i\epsilon$ stays positive, we can take its logarithm (with argument in $(-π, π)$, even in $(0, π)$ ). Anyhow we calculate in the distribution sense with a smooth function $g_4$, $$ \frac{d}{dy}\left[\ln(F(y)+i\epsilon)\right]=\frac{1+iy g_4(y)}{F(y)+i\epsilon}. \tag 2$$ With $\phi$ a test function (compactly supported smooth) we have by the Lebesgue Dominated Convergence Theorem, using $g_3\ge 0$ and $$ \frac{\vert iyg_4(y)\vert}{\vert y+i\epsilon+iy^2g_3(y)\vert}\lesssim \frac{\vert y\vert}{\vert y\vert+\epsilon}\le 1, $$ that $$ \lim_{\epsilon\rightarrow 0_+}\int \frac{iyg_4(y)}{y+i\epsilon+iy^2g_3(y)}\phi(y)dy= \int \frac{ig_4(y)}{1+iyg_3(y)}\phi(y), $$ implying from (2) that with a smooth function $\alpha$, we have $$ \alpha(y)+\frac{d}{dy}\left[\ln(F(y)+i\epsilon)\right]=\frac{1}{F(y)+i\epsilon}. $$ As a consequence, we find $$ \lim_{\epsilon\rightarrow 0_+}\langle \frac{1}{F(y)+i\epsilon},\phi(y)\rangle= \int \alpha(y) \phi(y) dy -\int \ln(F(y)+i0)\phi'(y) dy, $$ implying that the distribution $\frac{1}{F+i0}$ is well-defined and equal modulo a smooth function to the (distribution) derivative of $$ \ln\bigl(F(y)+i0\bigr)=\ln\bigl(y+iy^2 g_3(y)+i0)\bigr), \quad g_3\ge 0. \tag{3}$$ $[\mathbf 2]. $ $\bf\text{The wave-front-set.}$ Since the derivative is elliptic in one dimension, the wave-front-set of the derivative is the same as the wave-front-set of the initial distribution. We have \begin{multline} \ln\bigl(y+iy^2 g_3(y)+i\epsilon)\bigr)=\ln\vert y\vert+\text{smooth function}+i\arg\bigl(y+i(\epsilon+y^2g_3(y))\bigr) \\=\text{smooth function}+\ln\vert y\vert+iπ H(-y), \end{multline} where $H$ is the Heaviside function (indicatrix of $\mathbb R_+$). As a result, up to a smooth function, we have $$ \ln(F(y)+i0)=\ln\vert y\vert+iπ H(-y)\quad\text{with derivative}\quad (y+i0)^{-1}=\text{pv}(\frac 1y)-iπ\delta_0, $$ whose wave-front-set is known as $\{0\}\times (0,+\infty)$ (its Fourier transform is $-2iπ H$). The case $a<0$ reverses the arrows.

Answer 2

This question has already been answered but here is some additional material which might be of interest. One can show the existence of this distribution and even give a rather explicit representation, using only the following facts:

Every locally integrable function is a distribution;

Distributions can be differentiated and this operation preserves distributional convergence;

You can multiply a distribution by a smooth function.

We assume that $f$ is a smooth function as in your question which is such that $\log |f(x)|$ is locally integrable (and so a distribution) and whose derivative doesn´t vanish. We can then regard the reciprocal $\frac 1 f$ as a distribution by defining it to be $$\frac 1 {f´(x)} D\log|f(x)|$$ (distributional derivative). This is the principal value and its definition is simpler and more general than the usual ones. It includes the standard cases of inverse powers.

The condition on your function implies that the set of zeros of $f$ is discrete and we will simplify the representation by assuming that it has a single zero at the origin, is negative to the left and positive to the right thereof. There are standard techniques to accommodate to the case of a countable number of zeros.

We are motivated by the case where $f(x)=x$ which leads to the celebrated formula $$\lim_{\epsilon\to 0+}\frac 1 {x+i \epsilon}=\frac 1 x-i \pi \delta$$ where the first term on the right hand side is, as explained above, the principal value $D \log|x|$ and the second involves the Dirac distribution with singularity at the origin.

We now give the limit in your case which is obtained by looking at the distribution $\log (x+i \epsilon)$, taking the limit and then differentiating. The resulting formula is $$\lim_{\epsilon\to 0+}\frac 1 {f(x)+i \epsilon}=\frac 1 {f(x)}-i \pi \frac{\delta}{f´(0)}.$$ Again the first term on the right hand side is the principal value.

The limit around points at which $f$ is non zero can be calculated using classical methods.

One can write out a formula which is valid to the general case but this is rather tedious to do.