1$). Then \begin{equation} u(\vec{x})\ge C A^{1/p} |x|^{-(n-1)/p}, \end{equation} for all $x$ with $|x|\ge r_0$ and for some positive constant $C(n,p)$. \end{theorem} \begin{proof} Fix a particular radius $|\vec{x}|=R$ at which we verify \[u(\vec{x})\ge CA^{1/p}R^{-(n-1)/p}.\] We first claim that without loss of generality, $u$ is harmonic outside a compact set, and converges to zero as $|\vec{x}|$ goes to $\infty$. This is true because we can define (given any positive constants $\epsilon$ and $K$) the function \[\tilde{u}=\min \{u+\epsilon,K|\vec{x}|^{2-n}\}.\] Fix any $\epsilon>0$. Because $u$ is continuous, $u+\epsilon$ attains a maximum on the closed ball of radius $R$ around the origin. By choosing $K$ appropriately, we can ensure $KR^{2-n}$ is larger than $u+\epsilon$ on this ball. On the other hand, for $|\vec{x}|$ large enough, $\tilde{u}=K|\vec{x}|^{2-n}$ which is harmonic. The theorem, applied to this function $\tilde{u}$, would then prove that on the sphere $|\vec{x}|=R$, $u(\vec{x})+\epsilon\ge C$ for all $\epsilon>0$, and thus, $u(\vec{x})\ge C$. We can thus assume without loss of generality that $u$ is harmonic outside a compact set, and converges to zero as $|\vec{x}|$ goes to $\infty$. The consequence of this is that $f=\Delta u$, which is defined as a distribution, has compact support, and we can apply potential theory to prove that $u = f(\vec{x})*\frac{1}{\omega_n(n-2)}|\vec{x}|^{2-n}$. Since $u$ was superharmonic, $f$ is a positive measure. We next mollify $u$ so that it is smooth: let $\psi_j:\re^n\to\re$ be a sequence of spherically symmetric smooth functions supported in balls of radius $1/j$ around the origin, with $\int_{\re^n} \psi_j = 1$. Then $u_j=u*(\psi_j*\psi_j)$ is a smooth function. Since a continuous function defined on a compact set is uniformly continuous (see Rudin\cite{Rudin}, theorem 4.19), $u$ is uniformly continuous over compact subsets of $\re^n$. Therefore, as $j$ goes to infinity, $u_j$ converges to $u$ uniformly on compact sets. Furthermore, if we let $f_j=f*\psi_j$, and $g_j=\frac{1}{\omega_n(n-2)} |\vec{x}|^{2-n} *\psi_j$, then $u_j=f_j*g_j$, by associativity of convolutions. Since $u_j$ converges to $u$ uniformly on compact sets, we know that for $j$ large enough, \[\int_{S_R(0)}u_j^p\ge A/2.\] Thus if we can prove the theorem for $u_j$ smooth, then we will know that on the sphere $|\vec{x}|=R$, \begin{equation} u_j(\vec{x})\ge 2^{-1/p} C A^{1/p} |\vec{x}|^{-(n-1)/p}. \end{equation} Furthermore, by rescaling $\vec{x}$ by $1/R$ so that $R$ is replaced by $1$, we see that $A$ is replaced by $A/R^{n-1}$, so the theorem scales correctly. By dividing $u$ by $A^{1/p}R^{-(n-1)/p}$, we see that it suffices to prove the theorem when $A=1$. Thus, we need to prove the following: \begin{theorem} Suppose that $u:\re^n\to [0,\infty]$ for $n\ge 3$ is a smooth superharmonic function \[u=f*g\] with $f$ positive, continuous and of compact support, and $g(\vec{z})$ a spherically symmetric decreasing function of $|\vec{z}|$, and \begin{equation} \int_{S_r(0)} u(\vec{x})^p\ge 1 \end{equation} for $r\in [a,b]$, where $S_r(0)$ is the sphere of radius $r$ centered at the origin, $[a,b]$ is a closed interval in $(0,\infty)$, and $1

1$). Then
\begin{equation}
u(\vec{x})\ge C
\end{equation}
for all unit vectors $\vec{x}$, for
some positive constant $C(n,p,a,b)$.
\end{theorem}
Note that when we apply this, we will take $a=1$ and $b=2$.
By the spherical symmetry of the problem, it suffices to prove that
$u(-\vec{e}_1)\ge C$, where $\vec{e}_1$ is the basis vector in the $x_1$
direction.
The equation $u=f*g$ can be written
\[u(\vec{x})=\int_{\re^n}f(\vec{y})g(\vec{x}-\vec{y})\,d^n \vec{y}.\]
Let $\bar{f}(t)=\int_{S_t(0)}f(\vec{y})d^{n-1}\vec{y}$ for $t\ge 0$ and let
\[\bar{u}(\vec{x})=\int_0^\infty \bar{f}(t) g(\vec{x}-t\vec{e}_1)\,dt.\]
We view this procedure of moving from $f$ to $\bar{f}$ as that of concentrating
the source of $u$ to the positive $x_1$ axis. Then as the following lemma
will demonstrate, this will increase $\int_{S_r(0)}u^p$ but decrease
its value at $-\vec{e}_1$, thus reducing us to the axially symmetric case.
If $\vec{v}$ is a unit vector in $\re^n$, then we define a reflection
\[R_{\vec{v}}(\vec{x})
=\vec{x}-2\frac{\vec{v}\cdot\vec{x}}{|\vec{v}|^2}\vec{v}\]
as the reflection through the hyperplane perpendicular to $\vec{v}$.
Define the corresponding ``fold'' map:
\[\rho_{\vec{v}}(\vec{x})=\left\{\begin{array}{ll}
\vec{x},&\mbox{if $\vec{x}\cdot \vec{v}\ge 0$}\\
R_{\vec{v}}(\vec{x}),&\mbox{otherwise}
\end{array}\right.\]
If $f(\vec{x})$ is a real-valued function on $\re^n$, we will denote
by $f^{\wedge\vec{v}}$ (or $f^\wedge$ if $\vec{v}$ is understood), the
``folded'' function as follows:
\[f^{\wedge\vec{v}}(\vec{x})=\left\{\begin{array}{cc}
f(\vec{x})+f(R_{\vec{v}}(\vec{x})),& \vec{x}\cdot\vec{v}\ge 0\\
0,& \vec{x}\cdot\vec{v}<0
\end{array}\right.\]
For convenience we will define the following functions:
\[f^+(\vec{x})=\left\{\begin{array}{cc}
f(\vec{x}),& \vec{x}\cdot\vec{v}\ge 0\\
0,& \vec{x}\cdot\vec{v}<0
\end{array}\right.\]
\[f^-(\vec{x})=\left\{\begin{array}{cc}
f(\vec{x}),& \vec{x}\cdot\vec{v}< 0\\
0,& \vec{x}\cdot\vec{v}\ge 0
\end{array}\right.\]
so that
\[f(\vec{x})=f^+(\vec{x}) + f^-(\vec{x})\]
and
\[f^\wedge(\vec{x})=f^+(\vec{x})+f^-(R_v(\vec{x})).\]
\begin{lemma}
Let $\phi(\vec{x})$ be a bounded continuous real-valued function on
$\re^n$, spherically symmetric around the origin, and decreasing as a
function of $|\vec{x}|$. For any $f(\vec{x})$ a non-negative
integrable function on $\re^n$ with compact support, define $u_f=\phi
* f$ be the convolution of $\phi$ with $f$, and let
\[A_f=\int_{S^{n-1}} |u_f|^p\,d^{n-1}\omega\]
where $p\ge 1$.
Then
\[A_f \le A_{f^\wedge}.\]
for any unit vector $\vec{v}\in S^{n-1}$.
\end{lemma}
\begin{proof}
Let $\vec{x}\in \re^n$ be an arbitrary point with $\vec{x}\cdot
\vec{v}\ge 0$, and let $\vec{x}'=R_{\vec{v}}(\vec{x})$. Now
$u_{f^+}(\vec{x}')\le u_{f^+}(\vec{x})$ since $\phi$ is a decreasing
function of distance, and for any $\vec{y}$ with $\vec{x}\cdot
\vec{y}\ge 0$, we have that $\vec{x}$ is at least as close
to $\vec{y}$ as $\vec{x}'$ is.
Similarly, $u_{f^-}(\vec{x})\le u_{f^-}(\vec{x}')$, since
for any $\vec{y}$ with $\vec{x}\cdot\vec{y}<0$, we have that
$\vec{x}$ is at least as far from $\vec{y}$ as $\vec{x}'$ is.
If we let $a=u_{f^+}(\vec{x}')$, $b=u_{f^+}(\vec{x})$, $c=u_{f^-}(\vec{x})$,
$d=u_{f^-}(\vec{x}')$, then we have so far concluded that
$a\le b$ and $c\le d$. We will next prove that when $a\le b$ and $c\le d$,
\[(a+d)^p+(b+c)^p\le (a+c)^p+(b+d)^p.\]
First, without loss of generality, assume $b-a\ge d-c$. Then note that
$[a+c,b+d]$ and $[a+d,b+c]$ are nonempty intervals centered on the
same average point, but $[a+d,b+c]\subset [a+c,b+d]$, as can be
verified by comparing the lengths.
By convexity of the function $g(x)=x^p$ for $p\ge 1$, we have that the
graph of $x^p$ on $[a+c,b+d]$ lies below the secant line on the same
interval. Since $[a+d,b+c]\subset [a+c,b+d]$, the secant line for
$[a+d,b+c]$ lies below the secant line on $[a+c,b+d]$. Thus, the
midpoints of these secant line segments compare as follows:
\[\frac{(a+d)^p+(b+c)^p}{2} \le \frac{(a+c)^p+(b+d)^p}{2}\]
which gives us
\[(a+d)^p+(b+c)^p\le (a+c)^p+(b+d)^p\]
as claimed above.
Applied to our case, we have
\[\left(u_{f^+}(\vec{x}')+u_{f^-}(\vec{x}'))\right)^p
+\left(u_{f^+}(\vec{x})+u_{f^-}(\vec{x}))\right)^p
\le
\left(u_{f^+}(\vec{x}')+u_{f^-}(\vec{x}))\right)^p
+\left(u_{f^+}(\vec{x})+u_{f^-}(\vec{x}'))\right)^p
.\]
Since $f=f^++f^-$ and $u$ depends linearly on $f$,
$u=u_{f^+}+u_{f^-}$. Similarly, from $f^\wedge=f^++f^-\circ R_{\vec{v}}$, we
get $u=u_{f^+}+u_{f^-}\circ R_{\vec{v}}$. This gives us
\[(u_f(\vec{x}'))^p+(u_f(\vec{x}))^p\le
(u_{f^\wedge}(\vec{x}'))^p+(u_{f^\wedge}(\vec{x}))^p.\]
This holds whenever $\vec{x}\cdot\vec{v}\ge 0$, but by symmetry under
interchanging $\vec{x}$ and $\vec{x}'=R_{\vec{v}}(\vec{x})$ in the formula,
it also holds for all $\vec{x}\in S^{n-1}$.
Integrating this formula over all $\vec{x}\in S^{n-1}$ we get
\[\int_{S^{n-1}}(u_f(\vec{x}'))^p\,d^{n-1}\vec{x}
+\int_{S^{n-1}}(u_f(\vec{x}))^p\,d^{n-1}\vec{x}\le
\int_{S^{n-1}}(u_{f^\wedge}(\vec{x}'))^p\,d^{n-1}\vec{x}
+\int_{S^{n-1}}(u_{f^\wedge}(\vec{x}))^p\,d^{n-1}\vec{x}.\]
By changing variables from $\vec{x}$ to $\vec{x}'$ in the first and third
integrals, we obtain the second and fourth integrals, respectively, and,
dividing by two, we have
\[\int_{S^{n-1}}(u_f(\vec{x}))^p\,d^{n-1}\vec{x}\le
\int_{S^{n-1}}(u_{f^\wedge}(\vec{x}))^p\,d^{n-1}\vec{x}.\]
\end{proof}
\begin{lemma}
Let $\phi(\vec{x})$, $f(\vec{x})$, and $u_f$ be as in the previous lemma.
As before, define
\[A_f=\int_{S^{n-1}} |u_f|^p\,d^{n-1}\omega\]
where $p\ge 1$.
Similarly, define $\bar{f}$ as above:
\[\bar{f}(t)=\int_{S_t(0)}f(\vec{y})d^{n-1}\vec{y}\]
and
\[\bar{u}_f(\vec{x})=\int_0^\infty \frac{\bar{f}(t)}{\phi(\vec{x}-t\vec{e}_1)}
\,dt.\]
\[A_{\bar{f}}=\int_{S^{n-1}} |\bar{u}_f|^p\,d^{n-1}\omega\]
Then
\[A_f\le A_{\bar{f}}.\]
\end{lemma}
\begin{proof}
We will generate a sequence $f_n$, where $f_0=f$, and
\[A_{f_{n}}\le A_{f_{n+1}}.\]
This is obtained by choosing a sequence of unit vectors $\vec{v}_n$,
and setting $f_n=(f_{n-1})^{\wedge\vec{v}_{n-1}}$, applying the previous lemma.
The $\vec{v}_n$ will be chosen to make the domain of $f$ concentrate on the
positive $x_1$ axis. In order to keep track of this, we define inductively
the maximal possible domain of $f$ as follows:
Let $g_0(\vec{x})$ be the constant function 1, and recursively define
$g_n=(g_{n-1})^{\wedge\vec{v}_{n-1}}$. Then for each $n$, the support of
$f_n$ will be a subset of the support of $g_n$. Let
$W_n=supp(g_n)\cap S^{n-1}$. The $\vec{v}_n$ should be chosen so that
$W_n\supset W_{n+1}$, which means that the reflection
$\rho_{\vec{v}_n}$ should send $W_n$ into itself. We should choose
the $\vec{v}_n$ satisfying this restraint, in order to get a nested
sequence of $W_n$ that converges to the point $\vec{e}_1$.
Let $\alpha$ be an angle between $0$ and $\pi/2$. Consider the set of
vectors of the form
\[\vec{u}_{\alpha, j, \pm}=\sin\alpha\, \vec{e}_1 \pm \cos\alpha\, \vec{e}_j\]
where $j$ ranges over all the basis vectors that are not $\vec{e}_1$.
Let $C_\alpha$ be the ``hypercube'' set centered at $\vec{e}_1$:
\[C_\alpha=\{\vec{x}\in S^{n-1}|\vec{x}\cdot\vec{u}_{\alpha,j,\pm}\ge 0
\mbox{ for all $j$ and choices of $\pm$}\}\]
We take $\vec{v}_0=\vec{e}_1$, and thus $W_1$ is the right hemisphere
with $x_1\ge 0$. We then have that $W_1$ is the hypercube set $C_{\pi/2}$
centered at $\vec{e}_1$.
Inductively, whenever $W_k$ lies in the hypercube $C_{2\alpha}$ centered at
$\vec{e}_1$, we will choose $\vec{v}_{k+1}, ..., \vec{v}_{k+2n-2}$ to be
the $2n-2$ vectors $\vec{u}_{\alpha,j,\pm}$. We will then prove that
$W_{k+2n-2}$ lies in the hypercube $C_{\alpha}$ centered at $\vec{e}_1$.
It is clear that each fold map will make sure $\vec{x}\cdot\vec{u}_{\alpha,j,
\pm}\ge 0$ for the particular choice of $\alpha$, $j$, and $\pm$. But
we need to show that we are not at the same time making $\vec{x}\cdot
\vec{u}_{\beta,j',\pm}<0$ for some previous choice of $\beta$, $j'$, and $\pm$.
We will in fact show that the sequence of $W_m$'s is a nested sequence:
\[W_1\supset W_2\supset\cdots.\]
To verify this, suppose some $\vec{v}_{k+i}$ sent a vector $\vec{x}$
in some $W_{k+i}$ to something outside $W_{k+i}$. Since $W_{k+i}$ was
formed by ``folding'' using the $\vec{v}$'s, we know that for some
$\vec{v}_m$, we have that $\vec{x}\cdot\vec{v}_m\ge 0$ but
$\rho_{\vec{v}_{k+i}}(\vec{x})\cdot\vec{v}_m<0$. Furthermore, since
$\vec{x}$ was in $W_{k+i}$ to begin with, $\vec{x}\cdot\vec{v}_{k+i}<0$.
We have that $\vec{v}_{k+i}=\vec{u}_{\alpha,j_1,\pm}$ for some $j_1$ and
choice of $\pm$, and that $\vec{v}_m=\vec{u}_{\beta,j_2,\pm}$ for some
other similar choices. If $j_1\not=j_2$, then
$\vec{v}_{k+i}\cdot\vec{v}_m\ge 0$, and therefore
\begin{eqnarray*}
\rho_{\vec{v}_{k+i}}(\vec{x})\cdot\vec{v}_m
&=&R_{\vec{v}_{k+i}}(\vec{x})\cdot\vec{v}_m\\
&=&\left(\vec{x}-2\vec{v}_{k+i}\cdot\vec{x}\vec{v}_{k+i}\right)\cdot\vec{v}_m\\
&=&\vec{x}\cdot\vec{v}_m
-2(\vec{v}_{k+i}\cdot\vec{x})(\vec{v}_{k+i}\cdot\vec{v}_m)\\
&\ge&-2(\vec{v}_{k+i}\cdot\vec{x})(\vec{v}_{k+i}\cdot\vec{v}_m)\\
&\ge&0
\end{eqnarray*}
which gives the contradiction. We therefore consider $j_1=j_2$. If
the choices of $\pm$ for $\vec{v}_{k+i}$ and $\vec{v}_m$ are the same, then
again $\vec{v}_{k+i}\cdot\vec{v}_m\ge 0$ and by the same
argument arrive at the contradiction. So we know that the choices of
$\pm$ are different, and without loss of generality
\begin{eqnarray*}
\vec{v}_{k+i}&=&\vec{u}_{\alpha,j,+}=\sin\alpha \vec{e}_1+\cos\alpha\vec{e}_j\\
\vec{v}_m&=&\vec{u}_{\beta,j,-}=\sin\beta \vec{e}_1-\cos\beta\vec{e}_j.
\end{eqnarray*}
If we write $\vec{x}=a\vec{e}_1+b\vec{e}_j+\vec{y}$, then by induction
(that all the $W_k$ are in $W_1$) we know that $a\ge 0$. Furthermore,
since by assumption $\vec{x}\cdot\vec{v}_{k+i}<0$, we have that $b<0$.
Also since $\vec{x}\in W_k\subset C_{2\alpha}$, we know that
$a\sin(2\alpha)-b\cos(2\alpha)\ge 0$, which together with
$a^2+b^2+|\vec{y}|^2=1$, implies that $b\ge -\sin(2\alpha)$ and $a\ge
\cos(2\alpha)$. Then since $\vec{x}\cdot\vec{v}_{k+i}<0$,
\begin{eqnarray*}
\rho_{\vec{v}_{k+i}}(\vec{x})\cdot\vec{e}_j
&=&\left(\vec{x}-2\vec{v}_{k+i}\cdot\vec{x}\vec{v}_{k+i}\right)\cdot\vec{e}_j\\
&=&b-2(a\sin\alpha+b\cos\alpha)\cos\alpha\\
&\le&-\sin(2\alpha)
-2(\cos(2\alpha)\sin\alpha-\sin(2\alpha)\cos\alpha)\cos\alpha\\
&=&0
\end{eqnarray*}
and similarly
\begin{eqnarray*}
\rho_{\vec{v}_{k+i}}(\vec{x})\cdot\vec{e}_1
&=&\left(\vec{x}-2\vec{v}_{k+i}\cdot\vec{x}\vec{v}_{k+i}\right)\cdot\vec{e}_1\\
&=&a-2(\vec{v}_{k+i}\cdot\vec{x})\sin\alpha\\
&\ge&a\ge 0.
\end{eqnarray*}
Thus
\[\rho_{\vec{v}_{k+i}}(\vec{x})\cdot\vec{v}_m = \rho_{\vec{v}_{k+i}}
(\vec{x})\cdot(\sin(\beta)\,\vec{e}_1-\cos(\beta)\,\vec{e}_j) \ge 0,\]
which is, again, a contradiction.
In this way, we see that we have a nested sequence $W_k\supset
W_{k+1}$, each of which is contained in the hypercubes $C_{\alpha}$
centered at $\vec{e}_1$, with $\alpha$ decreasing to zero, so that
$\bigcap W_k=\{\vec{e}_1\}$, and $f_n$ converges in measure to a
delta function concentrated on the positive $x_1$ axis.
By the previous lemma, we have that
\[A_{f_n}\le A_{f_{n+1}}.\]
By the uniform convergence of $u_n$ to $\bar{u}$, we have that
$A_{f_n}$ converges to $A_{\bar{f}}$.
\end{proof}
Thus, if $\int_{S_r(0)}u^p$ is greater than or equal to one, then so
is $\int_{S_r(0)}\bar{u}^p$.
Also, note that $u(-\vec{e}_1)\ge \bar{u}(-\vec{e}_1)$, since
of all the points on $S_r(0)$, the point farthest away from $-\vec{e}_1$ is
$r\vec{e}_1$.
We have thus reduced our situation to the case where the source function
$f$ is supported on the positive $x$ axis, that is, using $\bar{f}$. So
we will have to show $\bar{u}(-\vec{e}_1)\ge C$.
To accomplish this, we interpret our inequality
\[\int_{S_r(0)}\bar{u}(x)^p\,d^{n-1}x\ge 1\]
which holds for $a\le r\le b$. Using the notation $\sigma_n$ for the
area of an $n$-dimensional sphere of radius 1, we have:
\begin{eqnarray*}
1&\le&\int_{S_r(0)}\bar{u}(x)^p\,d^{n-1}x\\
&=&\int_0^\pi \sigma_{n-2}(r\sin\theta)^{n-2}r\,d\theta\,\left(\bar{u}
(r\cos \theta e_1+r\sin\theta e_2)\right)^p\\
&=&\int_0^\pi \sigma_{n-2}r^{n-1}\sin^{n-2}\theta\,d\theta\,\left(
\int_0^\infty\frac{\bar{f}(t)\,dt}{|(r\cos\theta-t)e_1+r\sin\theta e_2|^{n-2}}
\right)^p\\
&=&\sigma_{n-2}r^{n-1}\int_0^\pi \sin^{-1+\epsilon}\theta \,d\theta\,\left(
\int_0^\infty\bar{f}(t) Q_{t,r}(\theta)\,dt\right)^p
\end{eqnarray*}
where
\[Q_{t,r}(\theta)=\frac{\sin^{\frac{n-2+1-\epsilon}{p}}\theta}{\left(
t^2+r^2-2tr\cos\theta\right)^{(n-2)/2}},\]
and $\epsilon>0$ is any positive constant.
Let
\[S_r(t)=\sup_{0\le\theta\le\pi}Q_{t,r}(\theta)\]
Then we have that
\[1\le \sigma_{n-2}r^{n-1}\int_0^\pi d\theta\left[\int_0^\infty \bar{f}(t)
S_r(t)\,dt\right]^p\]
so that
\[\int_0^\infty \bar{f}(t)S_r(t)\,dt
\ge \left(\pi\sigma_{n-2}r^{n-1}\right)^{-1/p}\]
for $a\le r\le b$.
Furthermore, since $k^{n-2}S_{kr}(kt)=S_r(t)$, setting $k=1/r$ gives
\[S_r(t)=\frac{1}{r^{n-2}}S_1(\frac{t}{r}).\]
Hence,
\[\int_0^\infty \bar{f}(t)\frac{1}{r^{n-2}}S_1(\frac{t}{r})\,dt\ge
\left(\pi\sigma_{n-2}r^{n-1}\right)^{-1/p}\]
so that
\[\int_0^\infty \bar{f}(t)\frac{1}{t}\frac{t}{r^2}S_1(\frac{t}{r})\,dt\ge
\left(\pi\sigma_{n-2}\right)^{-1/p}r^{(n-4)-(n-1)/p}\]
Now integrate with respect to $r$ from $a$ to $b$:
\[\int_0^\infty \bar{f}(t)\,dt\,\frac{1}{t}
\int_a^b \frac{t}{r^2}S_1(\frac{t}{r})\,dr\ge
\int_a^b \left(\pi\sigma_{n-2}\right)^{-1/p}r^{(n-4)-(n-1)/p}\,dr.\]
Let $y=t/r$, and substitute for $r$. Then
\begin{eqnarray*}
\int_0^\infty \bar{f}(t)\,dt\,\frac{1}{t}\int_{t/b}^{t/a} S_1(y)\,dy
&\ge&
\left(\pi\sigma_{n-2}\right)^{-1/p}\left(\frac{1}{(n-3)-(n-1)/p}\right)
\left.r^{(n-3)-(n-1)/p}\right|_a^b\\
&=&
\left(\pi\sigma_{n-2}\right)^{-1/p}\frac{1}{(n-3)-(n-1)/p}
\left(b^{(n-3)-(n-1)/p}-a^{(n-3)-(n-1)/p}\right).
\end{eqnarray*}
The possibility that $(n-3)-(n-1)/p=0$ is excluded since $p<(n-1)/(n-3)$.
Define
\[k=\sup_{0< t<\infty}\frac{t+1}{t}\int_{t/b}^{t/a} S_1(y)\,dy.\]
We will next prove that $k<\infty$. Then our theorem will be proved since
\begin{eqnarray*}
u(-\vec{e}_1)&\ge& \bar{u}(-\vec{e}_1)=\int_0^\infty \frac{\bar{f}(t)}{t+1}\,dt\\
&\ge&\frac{1}{k}\int_0^\infty\frac{\bar{f}(t)}{t+1}\,dt\,\frac{t+1}{t}
\int_{t/b}^{t/a} S_1(y)\,dy\\
&\ge&\frac{1}{k}\left(\pi\sigma_{n-2}\right)^{-1/p}\frac{1}{(n-3)-(n-1)/p}
\left(b^{(n-3)-(n-1)/p}-a^{(n-3)-(n-1)/p}\right)\\
&=&C.
\end{eqnarray*}
\begin{lemma}
The constant $k=\sup_{0\le t\le \infty}\frac{t+1}{t}\int_{t/b}^{t/a}S_1(y)\,dy$
is finite.
\end{lemma}
\begin{proof}
We will bound $S_1(y)$ from above. Recall that
\[S_1(t)=\sup_{0\le\theta\le\pi} Q_{t,1}(\theta)=
\sup_{0\le\theta\le\pi}\frac{(\sin\theta)^{(n-2+1-\epsilon)/p}}{\left(
t^2+1-2t\cos\theta\right)^{(n-2)/2}}.\]
Our first estimate involves the fact that by $\sin\theta\le 1$ and
$\cos\theta\le 1$, we have
\begin{equation}
S_1(t)\le \frac{1}{|t-1|^{n-2}}.\label{eqn:est1}
\end{equation}
Using this estimate, we see that
\[\frac{t+1}{t}\int_{t/b}^{t/a} S_1(y)\,dy\]
is continuous in $t$ in $\{0