Proof Let U be a nonempty, open, connected subset of $\mathbb {R}^{n}.$ A function $f:U\rightarrow \lbrack -\infty ,\infty )$ is said to be subharmonic if (1) f is not identically equal to $-\infty ,$ (2) f is upper semicontinuous, and (3) the average of f over a sphere centered at $x\in U$ is greater than or equal to $f(x),$ whenever the sphere is contained in $U.$ Such a function f is locally bounded above, because it is upper semicontinuous. Furthermore, f is in $L_{\mathrm {loc}}^{1}(U)$ and the distributional Laplacian of f is a non-negative distribution [Reference Hörmander22, Theorem 4.1.8]. If $f_{n}$ is a weakly decreasing sequence of subharmonic functions on U, the pointwise limit f of $f_{n}$ is easily seen to be subharmonic, provided f is not identically equal to $-\infty .$ (When computing the averages over spheres, apply monotone convergence to $f_{1}-f_{n}$ .) A smooth function $f:U\rightarrow \mathbb {R}$ is subharmonic if and only if the Laplacian of f is non-negative [Reference Azarin1, Theorem 2.6.4.2].

We now specialize to the case $n=2$ with $U=\mathbb {C}.$ The function $S(\lambda ,\varepsilon )$ is a smooth function of $\lambda $ for each $\varepsilon>0$ and the Laplacian of S with respect to $\lambda $ with $\varepsilon $ fixed is positive [Reference Mingo and Speicher24, Equation (11.8)]. Now, S can be computed as

$$\begin{align*}\int_{0}^{\infty}\log(\xi^{2}+\varepsilon)~d\mu_{\left\vert x-\lambda \right\vert }(\xi), \end{align*}$$

where $\mu _{\left \vert x-\lambda \right \vert }$ denotes the law (or spectral distribution) of $\left \vert x-\lambda \right \vert .$ After separating the log function into its positive and negative parts and applying monotone convergence to the negative part, we see that $s(\lambda )$ can be computed as

(2.6) $$ \begin{align} s(\lambda)=\int_{0}^{\infty}\log(\xi^{2})~d\mu_{\left\vert x-\lambda \right\vert }(\xi). \end{align} $$

Here, the integral of the positive part of the logarithm is finite but the integral of the negative part can be infinite, meaning that the integral is well defined but can equal $-\infty .$ If $\lambda $ is outside the spectrum of $x,$ then $\left \vert x-\lambda \right \vert $ is invertible, so the support of $\mu _{\left \vert x-\lambda \right \vert }$ of $\left \vert x-\lambda \right \vert $ does not include 0. In that case, the integral in (2.6) is finite. We conclude that s is not identically equal to $-\infty $ and is the decreasing limit of subharmonic functions; therefore, s is subharmonic.

We now apply Theorem 4.1.9 in [Reference Hörmander22] to the sequence $f_{n}(\lambda )=S(\lambda ,\varepsilon _{n}),$ for any decreasing sequence $\varepsilon _{n}$ of positive numbers tending to 0. We note that (1) $f_{n}(\lambda )$ is bounded above by $f_{1}(\lambda )$ and (2) when $\lambda $ is outside the spectrum of $x,$ the sequence $f_{n}(\lambda )$ is not tending to $-\infty $ . Then, [Reference Hörmander22, Theorem 4.1.9(a)] tells us that $f_{n}$ has a subsequence that converges in $L_{\mathrm {loc}}^{1}$ to some function $g.$ Then, this subsequence has a sub-subsequence converging pointwise almost everywhere to $g.$ But the whole sequence $f_{n}$ converges pointwise to $s,$ which means that $g=s$ almost everywhere. Finally, we apply a standard argument to the sequence $f_{n}$ in the metric space $L^{1}(K),$ for any compact subset K of $\mathbb {C}.$ Since every subsequence will have a convergent sub-subsequence and all the subsequential limits have the same value (namely, s), the entire sequence converges to s in $L^{1}(K).$ It is then easily seen that $S(\lambda ,\varepsilon )$ converges to s in $L^{1}(K)$ for every compact set $K.$

Proof The Hamiltonian is easily seen to be a constant of motion for any Hamiltonian system. If $\lambda _0=r_0e^{i\theta }$ , we compute

$$ \begin{align*} (r_0^2 - \varepsilon_0)p_{\varepsilon,0} - p_{\rho,0} &= \int_0^{\infty}\frac{-r^2_0 + 2\xi r_0\cos\theta - \varepsilon_0}{\xi^2 +r_0^2 -2\xi r_0\cos\theta +\varepsilon_0}\,d\mu(\xi)\\ &= -1 + p_2. \end{align*} $$

Then,

$$ \begin{align*}H_0 = -\varepsilon_0p_0(1 -1 + p_2) = -\varepsilon_0p_0p_2,\end{align*} $$

as claimed.

Proof We calculate

$$ \begin{align*} p_\rho\frac{d\rho}{dt} + p_\varepsilon\frac{d\varepsilon}{dt} &= p_\rho\frac{\partial H}{\partial p_\rho} + p_\varepsilon\frac{\partial H}{\partial p_\varepsilon} \\ &= \varepsilon p_\varepsilon - 2H = \varepsilon p_\varepsilon - 2H_0. \end{align*} $$

By Proposition 5.3 in [Reference Driver, Hall and Kemp10], we have

$$ \begin{align*}S(t,\lambda(t),\varepsilon(t)) = S(0,\lambda_0,\varepsilon_0) - H_0 t + \int_0^t \varepsilon(s)p_\varepsilon(s) \,ds.\end{align*} $$

Since

$$ \begin{align*}\frac{d\rho}{dt} = \frac{\partial H}{\partial p_\rho}= \varepsilon p_\varepsilon,\end{align*} $$

we have

(3.12) $$ \begin{align} \int_0^t \varepsilon(s)p_\varepsilon(s) \,ds = \log|\lambda(t)| - \log|\lambda_0|.\end{align} $$

Thus, we are done.

Proof It is an easy computation to show, using (3.4) and (3.5), that $d\phi /dt=0$ and that

$$ \begin{align*}\frac{d}{dt}(\varepsilon p_{\varepsilon}^2)=-\varepsilon p_{\varepsilon}^2(2\phi-1).\end{align*} $$

Then, since $\phi $ is a constant of motion, we obtain the claimed formula for $\varepsilon p_{\varepsilon }^2$ . Meanwhile, we can compute that

$$ \begin{align*} 2\phi(0)-1=\int_0^{\infty} \frac{2\varepsilon_0+(2r^2_0 - 2\xi r_0\cos(\theta))-(\xi^2 +r_0^2 -2\xi r_0\cos\theta +\varepsilon_0)}{\xi^2 +r_0^2 -2\xi r_0\cos\theta +\varepsilon_0}\,d\mu(\xi), \end{align*} $$

which simplifies to the claimed expression for C.

Proof If $\lambda _0$ is outside $\operatorname {supp}(\mu )$ and $\varepsilon _0$ is not too negative, the initial momenta in (3.6)–(3.8) will be well defined and the quantities $p_k$ in (3.10) will be well defined and positive. Specifically, we need $\varepsilon _0> -d(\lambda _0,\operatorname {supp}(\mu ))^2$ . In what follows, the statement “ $\varepsilon _0$ is slightly negative” will mean that $\varepsilon _0<0$ is chosen to be greater than $-d(\lambda _0,\operatorname {supp}(\mu ))^2$ and so that the quantity $\delta $ in (3.14) remains positive.

Recall that

$$\begin{align*}\frac{dp_{\varepsilon}}{dt}=-\frac{\partial H}{\partial\varepsilon}=-\frac {H}{\varepsilon}-\varepsilon p_{\varepsilon}^{2}. \end{align*}$$

Using Lemma 3.5, we can express $dp_{\varepsilon }/dt$ in a form that involves only $p_{\varepsilon }$ and constants of motion:

$$\begin{align*}\frac{dp_{\varepsilon}}{dt}=-\frac{H}{\varepsilon_{0}p_{0}^{2}}p_{\varepsilon }^{2}e^{Ct}-\varepsilon_{0}p_{0}^{2}e^{-Ct}. \end{align*}$$

Let $B=\varepsilon _{0}p_{0}^{2}$ and $y(t)=-\frac {H}{B}p_{\varepsilon } e^{Ct}+\frac {C}{2}$ . Since a Hamiltonian H is a constant of motion, we compute that, on the one hand,

$$\begin{align*}\frac{dy}{dt}=\frac{H^{2}}{B^{2}}p_{\varepsilon}^{2}e^{2Ct}-\frac{HC} {B}p_{\varepsilon}e^{Ct}+H, \end{align*}$$

but on the other hand,

$$\begin{align*}y^{2}=\frac{H^{2}}{B^{2}}p_{\varepsilon}^{2}e^{2Ct}-\frac{HC}{B} p_{\varepsilon}e^{Ct}+\frac{C^{2}}{4}. \end{align*}$$

We therefore find that

(3.15) $$ \begin{align} y^{2}-\frac{dy}{dt}=a^{2}, \end{align} $$

where

$$\begin{align*}a^{2}=\frac{C^{2}}{4}-H. \end{align*}$$

Now, (3.15) is a separable differential equation which can be solved as in Lemma 5.8 in [Reference Driver, Hall and Kemp10], as

(3.16) $$ \begin{align} y(t)=\frac{y_{0}\cosh(at)-a\sinh(at)}{\cosh(at)-y_{0}\frac{\sinh(at)}{a} }. \end{align} $$

We note that the right-hand side of (3.16) is an even function of $a,$ so that either of the square roots of $a^{2}$ may be used. Now, $y(t)$ (and therefore also $p_{\varepsilon }(t)$ ) will blow up when the denominator is zero, i.e., at $t=t_{\ast }$ where

(3.17) $$ \begin{align} t_{\ast}=\frac{1}{2a}\log\left( \frac{1+a/y_{0}}{1-a/y_{0}}\right). \end{align} $$

We now compute the value of $a^{2}$ as

(3.18) $$ \begin{align} a^{2} & =\frac{C^{2}}{4}-H\nonumber\\ & =\frac{p_{0}^{2}(r_{0}^{2}+\varepsilon_{0}-\frac{p_{2}}{p_{0}})^{2}} {4}+\varepsilon_{0}p_{0}p_{2}. \end{align} $$

We further simplify this result as

(3.19) $$ \begin{align} a^{2} & =\frac{p_{0}^{2}}{4}\left( \left( r_{0}^{2}+\varepsilon_{0} +\frac{p_{2}}{p_{0}}\right) ^{2}-4r_{0}^{2}\frac{p_{2}}{p_{0}}\right) \\ & =\frac{p_{0}p_{2}r_{0}^{2}}{4}\left( \delta^{2}-4\right) .\nonumber \end{align} $$

We then compute

$$ \begin{align*} y_{0} & =p_{2}+\frac{p_{0}r_{0}^{2}+p_{0}\varepsilon_{0}-p_{2}}{2}\\ & =\frac{p_{0}r_{0}^{2}+p_{0}\varepsilon_{0}+p_{2}}{2}\\ & =\frac{r_{0}\sqrt{p_{2}p_{0}}}{2}\delta. \end{align*} $$

We can then find the value of $p_{\varepsilon }(t)$ from the value of $y(t),$ with the result that

(3.20) $$ \begin{align} p_{\varepsilon}(t)=p_{0}e^{-Ct}\frac{\sqrt{\delta^{2}-4}\cosh(at)+(2r_{0} \sqrt{p_{0}/p_{2}}-\delta)\sinh(at)}{\sqrt{\delta^{2}-4}\cosh(at)-\delta \sinh(at)}. \end{align} $$

This expression is very similar to the one in [Reference Driver, Hall and Kemp10, Equation (5.45)], with the only difference being the factor of $\sqrt {p_{0}/p_{2}}$ in the second term in the numerator. (This factor does not appear in [Reference Driver, Hall and Kemp10] because $p_{2} =p_{0}$ when $x=1.$ ) We now claim that if $\varepsilon _{0}$ is either non-negative or slightly negative, the solution to the whole Hamiltonian system (3.4)–(3.5) $t_{\ast }$ when the denominator on the right-hand side of (3.21) becomes zero. A key step is to solve for $\varepsilon (t)$ in (3.13) as

(3.21) $$ \begin{align} \varepsilon(t)=\frac{1}{p_{\varepsilon}(t)^{2}}\varepsilon_{0}p_{0}^{2} e^{-Ct}. \end{align} $$

This formula is meaningful as long as $p_{\varepsilon }(t)$ remains nonzero.

We now claim that $p_{\varepsilon }(t)$ remains positive until it blows up, even if $\varepsilon _{0}$ is slightly negative. We consider first the case $\varepsilon _{0}\geq 0.$ In that case, by (3.18) and (3.19), $a^{2}\geq 0$ and $\delta \geq 4$ . We then claim that the coefficient of $\cosh (at)$ in the numerator of the right-hand side of (3.20) is at least as big as the absolute value of the coefficient of $\sinh (at).$ Thus, the numerator will be positive for all $t>0.$ To verify this claim, we compute that

$$\begin{align*}\delta^{2}-4-(2r_{0}\sqrt{p_{0}/p_{2}}-\delta)^{2}=\frac{4p_{0}\varepsilon _{0}}{p_{2}}\geq0. \end{align*}$$

We consider next the case in which $\varepsilon _{0}$ is slightly negative. From (3.18), we see that typically, $a^{2}$ remains positive even when $\varepsilon _{0}$ becomes slightly negative, in which case, the argument is as in the case $\epsilon _0\geq 0$ . But if $r_{0}^{2} =p_{2}/p_{0}$ at $\varepsilon _{0}=0,$ the value of $a^{2}$ – and thus the value of $\delta ^{2}-4$ – will become negative when $\varepsilon _{0}$ is slightly negative. In that case, we write $a=i\alpha ,$ where we can choose $\alpha>0.$ Then, the expression in (3.20) becomes

(3.22) $$ \begin{align} p_{\varepsilon}(t)=p_{0}e^{-Ct}\frac{\sqrt{4-\delta^{2}}\cos(\alpha t)-(\delta-2r_{0} \sqrt{p_{0}/p_{2}})\sin(\alpha t)}{\sqrt{4-\delta^{2}}\cos (\alpha t)-\delta\sin(\alpha t)}. \end{align} $$

The numerator on the right-hand side of (3.22) becomes zero at the time

$$ \begin{align*} t_1=\frac{1}{\alpha}\tan^{-1}\left( \frac{\sqrt{4-\delta^{2}} }{\delta-2r_{0}\sqrt{p_{0}/p_{2}}}\right) \end{align*} $$

while the denominator becomes zero at

$$ \begin{align*} t_2=\frac{1}{\alpha}\tan^{-1}\left( \frac{\sqrt{4-\delta^{2}} }{\delta}\right). \end{align*} $$

Let $\theta _1$ and $\theta _2$ denote the arguments of the inverse tangents in the formulas for $t_1$ and $t_2$ , respectively. Then, $\theta _2$ is positive, so that the value of the inverse tangent is in $(0, \pi /2).$ Now, $\theta _1$ could be positive and bigger than $\theta _2$ , in which case $t_1$ is bigger than $t_2$ . Alternatively, $\theta _1$ could be negative, in which case, to get a positive value of $t_1,$ the value of the inverse tangent must be bigger than $\pi /2.$ Either way, $t_1$ is greater than $t_2,$ showing that $p_{\varepsilon }(t)$ remains positive until it blows up.

Since $p_{\varepsilon }(t)$ remains positive until it blows up, the formula for $\varepsilon (t)$ in (3.21) remains nonsingular up to time $t_{\ast }.$ We can then follow the proof of [Reference Driver, Hall and Kemp10, Proposition 5.11] to construct a solution to the system (3.4)–(3.5) up to time $t_{\ast }$ .

Finally, by (3.21), if $\varepsilon _0>0$ , then $\varepsilon (t)$ will remain positive for $t<t_\ast $ .

Proof

Since $\lambda _0$ is assumed to be outside $\operatorname {supp}(\mu )$ , we may take $\varepsilon \rightarrow 0$ in (3.21) and $\varepsilon _0$ and $p_0$ will remain finite. Furthermore, as discussed in the proof of Theorem 3.6, as long as $\varepsilon _0$ is at most slightly negative, $p_\varepsilon (t)$ will remain positive for as long as the solution to the whole system exists. Thus, $\varepsilon (t)$ becomes identically zero in the limit, until the solution of the system ceases to exist. Meanwhile, when $\varepsilon _0$ approaches 0 (so that $\varepsilon (t)$ also approaches 0), we can see from (3.12) that $\vert \lambda (t)\vert $ approaches $\vert \lambda _0\vert $ . Since the argument of $\lambda (t)$ is a constant of motion, the proof is complete.

Proof Recall the definition of $\delta $ in Theorem 3.6. The key point is that when $\varepsilon _0=0$ , we have

$$\begin{align*}\delta^2-4=\frac{(\tilde{p}_0 r_0-\tilde{p}_2)^2}{\tilde{p}_0 \tilde{p}_2r_0^2}. \end{align*}$$

We then note that the quantity

$$\begin{align*}\frac{1}{\sqrt{\delta^2-4}}\log\left(\frac{\delta + \sqrt{\delta^2-4}}{\delta - \sqrt{\delta^2-4}}\right) \end{align*}$$

has the same value no matter which square root of $\delta ^2-4$ we use. It is therefore harmless to choose

$$\begin{align*}\sqrt{\delta^2-4}=\frac{\tilde{p}_0 r_0-\tilde{p}_2}{r_0 \sqrt{\tilde{p}_0 \tilde{p}_2}}. \end{align*}$$

In that case, we find that

$$\begin{align*}\lim_{\varepsilon_0 \rightarrow 0}\frac{\delta + \sqrt{\delta^2-4}}{\delta - \sqrt{\delta^2-4}} = \frac{\tilde{p}_0r_0^2}{\tilde{p}_2}\end{align*}$$

and

$$\begin{align*}\lim_{\varepsilon_0 \rightarrow 0}r_0\sqrt{p_0 p_2}\sqrt{\delta^2 -4} = \tilde{p}_0r_0^2 -\tilde{p}_2 \end{align*}$$

and the claimed formula holds, provided $\tilde {p}_0 r_0^2\neq \tilde {p}_2$ .

In the case $\tilde {p}_0 r_0^2= \tilde {p}_2$ , we have $\lim _{\varepsilon _0 \rightarrow 0} \delta = 2$ and, so that

$$ \begin{align*}\lim_{\varepsilon \rightarrow 0} \frac{1}{ \sqrt{\delta^2-4}}\log\left(\frac{\delta + \sqrt{\delta^2-4}}{\delta - \sqrt{\delta^2-4}}\right) = 1\end{align*} $$

and

$$ \begin{align*}\lim_{\varepsilon_0 \rightarrow 0} t_\ast = \frac{1}{\tilde{p}_2}\end{align*} $$

as claimed.

Finally, when $\lambda _0 = 0 \notin \operatorname {supp}(\mu )$ , we have that $\tilde {p}_0r_0^2 = 0< 1 =\tilde {p}_2.$ Then,

$$\begin{align*}\lim_{\varepsilon_0 \rightarrow 0}\frac{\delta + \sqrt{\delta^2-4}}{\delta - \sqrt{\delta^2-4}} = \infty \quad \text{and } \lim_{\varepsilon_0 \rightarrow 0}r_0p_0\frac{\sqrt{\tilde{p}_2}}{\sqrt{\tilde{p}_0}}\sqrt{\delta^2 -4} = 1 .\end{align*}$$

Thus,

$$\begin{align*}\lim_{\varepsilon \rightarrow 0} t_\ast(0,\varepsilon_0) = \infty = T(0),\end{align*}$$

completing the proof.

Proof We first claim that the function $\frac {\log (x) -\log (y)}{x-y}$ is decreasing in x with y fixed – and therefore, by symmetry, decreasing in y with x fixed. Taking the derivative with respect to x, we get

$$ \begin{align*} \frac{d }{dx}\left(\frac{\log(x) -\log(y)}{x-y}\right) &=\frac{(x-y)\frac{1}{x} + \log(\frac{y}{x})}{(x-y)^2} \\ &\leq \frac{(x-y)\frac{1}{x} +\frac{y}{x}-1}{(x-y)^2} \\ &=0, \end{align*} $$

where we have used the elementary inequality $\log (x) \leq x-1$ . Using this result, we find, in the first case, that

$$\begin{align*}0\leq \frac{\log(x_n) -\log(y_n)}{x_n-y_n} \leq\frac{\log(x_n) -\log(a)}{x_n-a}\rightarrow 0 \end{align*}$$

and, in the second case, that

$$\begin{align*}\frac{\log(x_n) -\log(y_n)}{x_n-y_n}\geq \frac{\log(b) -\log(y_n)}{b-y_n}\rightarrow +\infty, \end{align*}$$

as claimed.

Proof We write the quantities $\tilde {p}_k$ in (3.10) in polar coordinates as

$$\begin{align*}\int_0^{\infty} \frac{\xi^k}{r_0^2+\xi^2-2\xi r_0\cos(\theta_0)}\,d\mu(\xi). \end{align*}$$

Since $\tilde {p}_k$ is an even function of $\theta _0$ , so is $T(r_0 e^{i\theta _0})$ . We therefore consider only the limit at $\theta _0$ approaches 0 from above. We then note that $\theta _0$ decreases toward zero, and the value of $\tilde {p}_k$ increases. Thus, by the monotone convergence theorem,

(3.24) $$ \begin{align} \lim_{\theta_0\rightarrow 0}\tilde{p}_k=\int_0^{\infty}\frac{\xi^k}{(\xi-r_0)^2}\,d\mu(\xi). \end{align} $$

Point (1) of the proposition is then clear in the case that $\tilde {p}_0$ and $\tilde {p}_2$ are finite at $\theta _0=0$ .

We then consider the case when at least one of $\tilde {p}_0$ and $\tilde {p}_2$ is infinite at $\theta _0=0$ , in which case, $\tilde {p}_0$ must be infinite. Now, since $\mu \neq \delta _0$ , the value of $\tilde {p}_2$ at $\theta _0 = 0$ is not zero. Furthermore, the value of $\tilde {p}_2$ at any $\theta _0$ is bounded below by its (positive) value at $\theta _0=0$ . Thus, by (3.24) and Lemma 3.9, the limit of $T(r_0 e^{i\theta _0})$ exists and is zero. We conclude that the limit in Point (1) of the proposition exists for all nonzero $r_0$ and that this limit is zero whenever the value of $\tilde {p}_0$ at $\theta _0=0$ is infinite. But by Lemma 3.8, $\tilde {p}_0=\infty $ (at $\theta _0=0$ ) for $\mu $ -almost every nonzero value of $r_0$ .

Finally, if $\lambda _0\rightarrow \infty $ , then the quantities $\tilde p_0$ and $\tilde p_2$ in (3.23) tend to zero, so that by the second part of Lemma 3.9, $T(\lambda _0)$ tends to $+\infty $ .

Proof By Point (2) of Proposition 3.10, the closed set $\overline \Sigma _t$ contains $\mu $ -almost every nonzero real number.

Proof We use the subscript notation $f_\theta $ for the partial derivative of a function f in the angular direction. Since $T(\overline {\lambda _0})=T(\lambda _0)$ , we see that $T(r_0 e^{i\theta })$ is an even function of $\theta $ . It therefore suffices to show that $T_\theta $ is positive when $\Im (\lambda _0)$ is positive. Let

$$ \begin{align*}R = \frac{\tilde{p}_0r_0^2}{\tilde{p}_2}.\end{align*} $$

Note that T can be computed as

$$ \begin{align*}T(r_0e^{i\theta}) = \frac{1}{\tilde{p}_2(R-1)}\log(R).\end{align*} $$

We first make a preliminary calculation:

$$ \begin{align*} T_\theta = \left(\frac{1}{\tilde{p}_2 (R-1)}\right)_\theta \log(R)+\frac{1}{\tilde{p}_2(R-1)}\frac{R_\theta}{R}. \end{align*} $$

We then use the inequality

(3.26) $$ \begin{align}1-1/x\leq \log x \leq x-1,\end{align} $$

which may be proved by writing $\log x$ as the integral of $1/y$ from 1 to x and then bounding $1/y$ between 1 and $1/x$ (for $x<1$ ) and between $1/x$ and 1 (for $x>1$ ).

In the case that $\left (\frac {1}{\tilde {p}_2 (R-1)}\right )_\theta $ is positive, we use the first inequality in (3.26) to give

$$ \begin{align*} T_\theta\geq \left(\frac{1}{\tilde{p}_2 (R-1)}\right)_\theta (1-1/R)+\frac{1}{\tilde{p}_2(R-1)}\frac{R_\theta}{R}. \end{align*} $$

Simplifying this result gives

$$ \begin{align*} T_\theta \geq -\frac{(\tilde{p}_2)_\theta}{\tilde{p}_2^2R}. \end{align*} $$

But

$$ \begin{align*} (\tilde{p}_2)_\theta = -\Im (\lambda_0)\int_0^\infty \frac{\xi^3}{(\xi^2 -2\xi r_0\cos \theta + r_0^2)^2} \,d\mu(\xi). \end{align*} $$

Thus, $T_\theta $ is positive when $\Im (\lambda _0)$ is positive and we are done in this case.

In the case that $\left (\frac {1}{\tilde {p}_2 (R-1)}\right )_\theta $ is negative, we use the second inequality in (3.26) to give

$$ \begin{align*} T_\theta\geq \left(\frac{1}{\tilde{p}_2 (R-1)}\right)_\theta (R-1)+\frac{1}{\tilde{p}_2(R-1)}\frac{R_\theta}{R}. \end{align*} $$

Simplifying this result gives

$$ \begin{align*} T_\theta &\geq - \frac{[(\tilde{p}_2)_\theta R+\tilde{p}_2R_\theta]}{\tilde{p}_2^2 R} \\ &= -\frac{(\tilde{p}_2R)_\theta}{\tilde{p}_2^2R}. \end{align*} $$

But since $\tilde {p}_2R = \tilde {p}_0r_0^2$ , we compute that

$$ \begin{align*} [\tilde{p}_2R]_\theta = -\Im (\lambda_0)\int_0^\infty \frac{r_0^2\xi}{(\xi^2 -2\xi r_0\cos \theta + r_0^2)^2} \,d\mu(\xi). \end{align*} $$

Thus, $T_\theta $ is positive when $\Im (\lambda _0)$ is positive and we are done in this case.

Proof of Proposition 3.13.

Suppose that $\lambda _0 =r_0e^{i\theta _0}$ with $\theta _0 \neq 0$ . Then, by the monotonicity of $T(r_0e^{i\theta _0})$ with respect to $\theta _0$ (Lemma 3.14), $|\theta _0| < \theta _t(r_0)$ if and only if $T(\lambda _0) <t$ . here

Proof The function T in Definition 3.1 is continuous outside $\operatorname {supp}(\mu )$ and, in particular, outside $[0, \infty )$ .

First, assume $\theta _t(r_0) = \pi ^+$ . Note that by monotonicity of $T(r_0e^{i\theta _0})$ with respect to $\theta _0$ (Lemma 3.14), $\theta _t(r_0) = \pi ^+$ if and only if $T(r_0e^{i\pi }) <t$ . By the continuity of T, there exists a neighborhood B of $r_0$ such that $T(s e^{i\pi }) <t$ for all $s \in B_1$ . Hence, $\theta _t(s) = \pi ^+$ for all $s \in B$ .

Next, assume $\theta _t(r_0) = \pi $ . Let $\varepsilon>0$ , where it is harmless to assume $\varepsilon <\pi $ . Then, by the definition of $\theta _t$ , we must have $T(r_0 e^{i(\pi -\varepsilon )})<t$ . Thus, by the continuity of T, there is some neighborhood B of $r_0$ such that $T(s e^{i(\pi -\varepsilon )})<t$ for all $s\in B$ . Finally, by the monotonicity of T, we must have $\theta _t(s)>\pi -\varepsilon $ for $s\in B$ , meaning that either $\theta _t(s)\in (\pi -\varepsilon ,\pi ]$ or $\theta _t(s)=\pi ^+$ .

Finally, assume $\theta _t(r_0) < \pi .$ Let $\varepsilon>0$ , where it is harmless to assume that $\theta _t(r_0)+\varepsilon <\pi $ and (in the case $\theta _t(r_0)>0$ ) that $\theta _t(r_0)-\varepsilon>0$ . Then, by monotonicity of T, we have $T(r_0e^{i(\theta _t(r_0)+\varepsilon )})> t$ . Since T is continuous, there exists a neighborhood U of $r_0$ such that

$$ \begin{align*}T(s e^{i(\theta_t(r_0)+\varepsilon)})> t, \quad \forall s \in U.\end{align*} $$

Hence, $\theta _t(s) \leq \theta _t(r_0) + \varepsilon $ , for all $s \in U.$

In the case $\theta _t(r_0) =0$ , we may then take $B=U$ . In the case $\theta _t(r_0) \neq 0$ , the monotonicity of T shows that $T(r_0 e^{i(\theta _0(r_0)-\varepsilon )}) < t.$ Since T is continuous, there exists a neighborhood V of $r_0$ such that

$$ \begin{align*}T(s e^{i(\theta_0(r_0)-\varepsilon)}) < t, \quad \forall s \in V.\end{align*} $$

Hence, $\theta _t(s) \geq \theta _t(r_0) - \varepsilon $ , for all $s \in V.$ Thus, we may take $B=U\cap V$ .

Proof of Proposition 3.15.

By Proposition 3.13, $\Sigma _t$ is the set $r_0 e^{i\theta _0}$ with $r_0\neq 0$ and $\vert \theta _0\vert <\theta _t(r_0)$ . Fix $r_0e^{i\theta _0}\in \Sigma _t$ . If $\theta _t(r_0)=\pi ^+$ , then by Point (1) of Lemma 3.16, there is a neighborhood B of $r_0$ on which $\theta _t=\pi ^+$ , in which case all $s e^{i\theta }$ with $s\in B$ belong to  $\Sigma _t$ .

If $\theta _t(r_0)\neq \pi ^+$ , set $\varepsilon =(\theta _t(r_0)-\vert \theta _0\vert )/2$ . Then, by Points (2) and (3) of Lemma 3.16, there is a neighborhood B of $r_0$ on which

$$ \begin{align*}\theta_t(s)>\theta_t(r_0)-\varepsilon=\theta_0+\varepsilon.\end{align*} $$

Then, all $s e^{i\theta }$ with $s\in B$ and $\theta $ within $\varepsilon $ of $\theta _0$ are in $\Sigma _t$ .

Proof Fix $\lambda _0=r_0 e^{i\theta _0}$ , with $-\pi <\theta \leq \pi $ , belonging to $\partial \Sigma _t\setminus \{0\}$ Then, $\lambda _0$ cannot be in the open set $\Sigma _t$ . Since $T(\bar \lambda _0)=T(\lambda _0)$ , we can assume that $\lambda _0$ is in the closed upper half plane. Then, $\theta _t(r_0)$ cannot be $\pi ^+$ or $\lambda _0$ would be in $\Sigma _t$ . If $\theta _0=0$ , then $\theta _t(r_0)$ must be zero, or $\lambda _0$ would be in $\Sigma _t$ . If $0<\theta _0\leq \pi $ , then $\theta _0$ cannot be less then $\theta _t(r_0)$ or $\lambda _0$ would be in $\Sigma _t$ . But also $\theta _0$ cannot be greater than $\theta _t(r_0)$ or $\theta _t(r_0)$ would be less than $\pi $ – and then by the continuity of $\theta _t$ , there would be a neighborhood of $\lambda _0$ outside $\Sigma _t$ . In that case, $\lambda _0$ could not be in $\partial \Sigma _t$ . Thus, in all cases, $\theta _0=\theta _t(r_0)$ , establishing the first part of the proposition.

For the second part, $T(\lambda _0)$ cannot be less than t or $\lambda _0$ would be in $\Sigma _t$ . For the third part, the assumptions ensure that $\lambda _0$ is not in $\operatorname {supp}(\mu )$ , in which case, T is continuous at $\lambda _0$ . Then, since $\lambda _0$ is a limit of points with $T<t$ , we have $T(\lambda _0)\leq t$ (and also $T(\lambda _0)\geq t$ ).

Lastly, let $\lambda _0$ be such that $T(\lambda _0) =t$ and $\arg (\lambda _0) \neq 0$ . Then, $\lambda _0$ is, by definition, not in $\Sigma _t$ . But $\lambda _0$ must be in the closure of $\Sigma _t$ , by the monotonicity of T with respect to $\theta $ . Thus, $\lambda _0$ must be in the boundary of $\Sigma _t$ .

Proof (Similar to Lemma 6.3 in [Reference Driver, Hall and Kemp10]) Note that if $\varepsilon _0 = 0$ , then $\varepsilon (t) \equiv 0$ and $\lambda (t) \equiv \lambda _0$ . Thus, $U_t(\lambda _0,0) = (\lambda _0,0)$ . Then, we only need to show that $\frac {\partial \varepsilon }{\partial \varepsilon _0}(t;\lambda _0,0)> 0$ . From Lemma 3.5, we have

$$ \begin{align*} \frac{\partial \varepsilon}{\partial \varepsilon_0}(t;\lambda_0,0) &= \frac{1}{p_\varepsilon(t)^2} p_0^2e^{-Ct} + \varepsilon_0 \frac{\partial}{\partial \varepsilon_0}\left[\frac{1}{p_\varepsilon(t)^2} p_0^2e^{-Ct}\right]\bigg\vert_{\varepsilon_0 = 0}\\ &=\frac{1}{p_\varepsilon(t)^2} p_0^2e^{-Ct}>0, \end{align*} $$

as claimed.

Proof

We first establish the result for $\lambda \notin \overline {\Sigma }_t \cup [0, \infty )$ . We take $\lambda _0=\lambda $ in Lemma 3.20. Then, by the inverse function theorem, the map $U_t$ in (3.28) has a local inverse near $U_t(\lambda ,0)=(\lambda ,0)$ .

Now, the inverse of the matrix in Lemma 3.20 will have a positive entry in the bottom right corner; that is, $U_t^{-1}$ has the property that $\partial \varepsilon _0/\partial \varepsilon>0$ . Thus, the $\varepsilon _0$ -component of $U_t^{-1}(\lambda ,\varepsilon )$ will be positive for $\varepsilon $ small and positive. In that case, the solution of the Hamiltonian system (3.4)–(3.5) will have $\varepsilon (u)>0$ up to the blow-up time. In turn, the blow-up time with initial conditions $(\lambda _0,\varepsilon _0)=U_t^{-1}(\lambda ,\varepsilon )$ will exceed t for $\varepsilon $ close to 0.

We now let “HJ” denote the quantity on the right-hand side of the Hamilton–Jacobi formula (3.27):

(3.29) $$ \begin{align} \mathrm{HJ}(t,\lambda_0,\varepsilon_0) = S(0,\lambda_0,\varepsilon_0) + \varepsilon_0p_0p_2t +\log|\lambda(t)| - \log|\lambda_0| .\end{align} $$

If $\varepsilon $ is small and positive, the Hamilton–Jacobi formula (3.27) tells us that

(3.30) $$ \begin{align} S(t,\lambda,\varepsilon) = \mathrm{HJ}(t,U_t^{-1}(\lambda,\varepsilon)).\end{align} $$

Now, the Hamilton–Jacobi formula is not directly applicable when $\varepsilon =0$ , because $S(t,\lambda ,\varepsilon )$ is defined only for $\varepsilon>0$ . But the map $U_t(\lambda _0,\varepsilon _0)$ , defined in terms of the Hamiltonian system (3.4)–(3.5), does makes sense when $\varepsilon <0$ . Thus, the right-hand side of (3.30) does make sense when $\varepsilon $ is zero or even slightly negative. Thus, (3.30) provides a way of computing the limit of $S(t,\lambda ,\varepsilon )$ as $\varepsilon $ approaches zero.

In the limit $\varepsilon _0 \rightarrow 0^+$ with $\lambda $ fixed, the inverse function theorem tells us that $U_t^{-1}(\lambda ,\varepsilon ) \rightarrow (\lambda ,0)$ . Thus, the limit of $S(t,\lambda ,\varepsilon )$ as $\varepsilon $ tends to zero from above can be computed by taking $\lambda _0=\lambda $ and $\varepsilon _0=0$ on the right-hand side of (3.29). Since $\lambda (t)$ becomes zero in this limit, as discussed at the beginning of this section, we get

$$\begin{align*}\lim_{\varepsilon\rightarrow 0^+}S(t,\lambda,\varepsilon)=\lim_{\varepsilon_0\rightarrow 0^+,\,\lambda_0\rightarrow\lambda}S(0,\lambda_0,\varepsilon_0)= \int_0^\infty\log(|\xi-\lambda|^2)\,d\mu(\xi), \end{align*}$$

where the second term on the right-hand side of (3.29) tends to zero as $\varepsilon _0$ tends to zero, because $p_0$ and $p_2$ remain finite when $\lambda _0$ is outside $\operatorname {supp}(\mu )$ . The limit of $S(t,\lambda ,\varepsilon )$ as $\varepsilon $ tends to zero from above can be computed by putting $\lambda (t;\lambda _0,\varepsilon _0)= \lambda $ and letting $\varepsilon _0 \rightarrow 0$ and $\lambda _0 \rightarrow \lambda $ on the right-hand side of (3.30). We therefore obtain the desired result when $\lambda $ is outside $\overline \Sigma _t\cup [0, \infty )$ .

We now have that

(3.31) $$ \begin{align} s_t(\lambda) = \int_0^{\infty}\log(|\xi-\lambda|^2)\,d\mu(\xi) \end{align} $$

holds for any $\lambda \in \mathbb {C}$ outside $\overline {\Sigma }_t \cup [0, \infty )$ . Since $s_t(\lambda )$ is subharmonic, it is locally integrable everywhere (see Proposition 2.2). Thus, $s_t$ can be interpreted as distribution by integrating against test functions with respect to Lebesgue measure on the plane. It follows that when computing $s_t$ in the distribution sense, we can ignore sets of Lebesgue measure zero, such as $[0, \infty )$ .

We conclude, then, that the formula (3.31) continues to hold in the distribution sense on the complement of $\overline {\Sigma }_t$ . Now, the right-hand side of (3.31) is a smooth function outside $\overline \Sigma _t\cup \{0\}$ , by Lemma 3.12. Its Laplacian in the distribution sense is then the Laplacian in the classical sense, which can be computed by putting the Laplacian inside the integral, giving an answer of 0. Thus, $\mu _t$ is zero outside $\overline \Sigma _t\cup \{0\}$ . Once this result is known, it is easy to see that (3.31) actually holds in the pointwise sense outside $\overline \Sigma _t\cup \{0\}$ .

Finally, we compute the mass of $\mu _t$ at the origin, in the case that $0$ is outside $\overline \Sigma _t$ . In that case, we can write

(3.32) $$ \begin{align} s_t(\lambda) = \log(|\lambda|^2)\mu(\{0\}) + \int_{\operatorname{supp}(\mu)\setminus\{0\}} \log(|\xi-\lambda|^2)\,d\mu(\xi), \end{align} $$

where $\operatorname {supp}(\mu )\setminus \{0\}$ is contained in $\overline \Sigma _t$ by Lemma 3.12. Thus, the second term on the right-hand side of (3.32) is harmonic outside $\overline \Sigma _t$ . Thus, if $0 \notin \overline {\Sigma }_t$ , we obtain

$$\begin{align*}\frac{1}{4\pi}\Delta s_t = \mu(\{0\})\delta_0 \quad \text{on } \overline{\Sigma}_t^c,\end{align*}$$

as claimed.

Proof Assume 0 is in $\operatorname {supp}(\mu )$ but $(0, c)$ is outside $\operatorname {supp}(\mu )$ for some $c>0$ . Then, for $\lambda \notin \operatorname {supp}(\mu )$ , we have

$$\begin{align*}\int_0^\infty \frac{\xi+\lambda}{\xi-\lambda}\,d\mu(\xi)=-\mu(\{0\})+\int_c^\infty \frac{\xi+\lambda}{\xi-\lambda}\,d\mu(\xi) \end{align*}$$

and the result follows easily.

Proof Let $\lambda _0 \notin \overline {\Sigma }_s$ be nonzero. Then, $p_{\varepsilon ,0}$ is the same as $p_0$ in (3.10) in Section 3. Thus, by Corollary 3.12, $p_{\varepsilon ,0}$ remains finite as $\varepsilon _0 \rightarrow 0$ . Meanwhile, as $\varepsilon _0 \rightarrow 0$ , we have

$$ \begin{align*} p_{\lambda,0} &= \frac{\partial}{\partial \lambda_0} S(s,s,\lambda_0,\varepsilon_0)\\ &= \frac{\partial}{\partial \lambda_0} \int_0^{\infty} \log\left(|\xi-\lambda_0|^2\right)d\mu(\xi) \\ &= \int_0^{\infty} \frac{1}{\lambda_0 -\xi} \,d\mu(\xi). \end{align*} $$

It follows that

$$ \begin{align*} \lim_{\varepsilon_0 \rightarrow 0} 2\lambda_0p_{\lambda,0} -1 &= 2\lambda_0 \int_0^{\infty} \frac{1}{\lambda_0 -\xi} \,d\mu(\xi) -1\\ &= - \int_0^{\infty} \frac{\xi+\lambda_0}{\xi-\lambda_0} \,d\mu(\xi). \end{align*} $$

Hence, letting $\varepsilon $ tend to zero in the formula $({4.6})$ for $\lambda (\tau )$ , we get

$$ \begin{align*} \lim_{\varepsilon_0 \rightarrow 0}\lambda(\tau) &= \lim_{\varepsilon_0 \rightarrow 0} \lambda_0\exp\left\{\frac{\tau-s}{2}\left(\varepsilon_0p_{\varepsilon,0} + 2\lambda_0p_{\lambda,0} - 1\right)\right\}\\ &= \lambda_0\exp\left[\frac{\tau-s}{2}\left(-\int_0^{\infty} \frac{\xi+\lambda_0}{\xi-\lambda_0} \,d\mu(\xi)\right)\right] \\ &= f_{s-\tau}(\lambda_0). \end{align*} $$

The same calculation is applicable if $\lambda _0\notin \overline \Sigma _s$ equals 0, provided that 0 is not in the support of $\mu $ .