Let us come now to the second exercise.
The second exercise concerns the superharmonic functions, which is a generalization of harmonics
function.
And of course there are superharmonic functions in which the inequality here is reversed.
By the way, the inequality is written with the minus Laplacian because the minus Laplacian
is a positive operator, as we shall see.
And we'll do some spectral analysis, maybe in the following of the course.
And therefore, or simply by Fourier transform.
And therefore, superharmonic means that applying the minus Laplacian operator, you should get
something negative instead of superharmonic, something positive.
So we say that Vc2 of a sum opposite u is superharmonic if minus Laplacian of V is less
equal than zero, which means Laplacian of V is larger equal than zero.
Let's try to prove that a superharmonic function, so this is why not the mean value property
for the averages on ball, but an inequality, which means V of x is less equal than the
average of the ball Vxr of V in the y.
For every x in u and r positive, such that the Vxr is inside u or r less than, I mean,
the distance from x to the boundary of u, if you prefer.
So since u is open, you can always find the rx small enough, such that Vxr is inside u.
So for all r smaller than rx, we define this function phi r, which is the integral of the
average integral of Vxr of V of y in the y.
That is one over omega nr2dn integral of Vxr, V of y dy.
Now we change variables, y equal to x plus rz, so that we get one over omega nr2dn integral
over V01, V of x plus rz, r to the end is z, and so we simplify r.
And so we see that the dependence on r is actually only inside V, and this is very nice,
because what we want to show is that phi of r converges to V of x as r goes to zero plus,
and then phi is non-decreasing.
This will tell us that phi of zero plus is less equal than phi of r for every r positive,
phi of r being the average of V over the ball Vxr, and phi of zero plus being V of x.
How to prove this thing follows by the continuity of V, because, in a very particular situation,
it means it's continuous inside u.
Because for every epsilon positive, there exists delta positive, such that for all r
and zero delta, the absolute value of V of x plus rz minus V of x is less than epsilon,
x is fixed.
Therefore, we simply subtract V of x from one over omega n integral over V01, V of x
plus rz, we bring it inside, because it's of course an average integral, we bring inside
the absolute value, and this becomes less than epsilon, and so well done.
Now we need to prove that phi is non-decreasing, so it's Vc2, then V in u, then Vc2 up to the
boundary of the ball Bxr, because the ball Bxr is containing the other ball Bxrx, which
is containing u, and therefore, all the closure of this ball is inside u, because there is
a positive distance, basically, from the ball to the boundary of u.
Hence, as a simple consequence of Lebesgue Dominator Convergence Theorem, we get that
phi is a C1 function over the little interval zero rx, and the first derivative is given
by the constant, one over the measure of the ball, integral over the ball of the gradient
of V evaluated in x plus ry, scalar product with y in dy.
Now, notice that the gradient of norm of y squared over 2 is actually y, so the trick
to evaluate this term is to apply integration of a path formula in a clever way, and twice.
So the integral of a B01 of dv evaluated in x plus ry times y is equal to the integral
of the ball, dv evaluated in x plus ry times scalar product, the gradient of the norm of
y squared over 2.
Now we apply the Gauss Green formula, because if you wanted one of the Green's identity,
because here we have gradient against gradient, so minus, we go to minus, the integral over
the ball of, so we take away this gradient, which becomes the divergence of this, so y
squared over 2 times the divergence of the gradient is laplacian of V evaluated in x
plus ry times r, because y is multiplied by r, so chain rule.
Notice the integral over the boundary of the unit ball, the gradient of V evaluated
in x plus ry times norm of y squared over 2 times minus y over norm of y, because this
is the unit interior normal to the ball.
Of course, minus and minus is plus if you want, in the h and minus 1, or in the surface
measure as you prefer.
But now recall that the norm of y is equal to 1 on the ball, on the boundary of the ball,
of the unit ball.
I mean, of course.
So actually this norm of y squared here is actually 1 half.
And this is very nice, because I got rid of this term.
Maybe at first I make it appear and here it disappears, and instead here it remains, norm
of y squared, r over 2, that's all nice.
Now what do we have in the second integral?
We have the integral over a boundary, the boundary of B01, a constant 1 alpha, the gradient
of a function against the gradient of V evaluated in x plus ry against the unit interior normal
to the unit ball.
But this, by Green's identity, is equal to the integral over B01 of r over 2, the Laplacian
of V evaluated in x plus ry.
So the effect of r comes from the fact that I'm taking the divergence.
So maybe this equality is easier to see in the reverse.
The integral over B01 of the divergence of the gradient of V is equal to minus the integral
over the boundary of B01 of dV against the unit interior normal.
And there is another factor, I mean for effective, if I take the divergence of the gradient of
V evaluated in x plus ry, chain rule.
So we arrive over here, and now we are very happy, because we have Laplacian of V, Laplacian
of V, integral over the same thing.
This is r over 2 times the integral over B01, 1 minus normal y squared times the Laplacian
of V.
Now, this is non-negative, because the Laplacian, I'm sorry, because normal y is less than 1.
Laplacian of V is non-negative by assumption, because u is supermonic, V is supermonic.
And therefore, phi prime is non-negative.
Therefore, phi is non-decreasing.
And so phi of rho is less equal than phi of r for all rho and r, rho less than r, between
0 and rx.
So V of x is the limit, as we show, of phi rho as rho goes to 0 plus, is less equal than
phi of r, which is the average integral over Bxr of V of y and dy.
Some of you, I noticed this with the, by defining phi as the boundary integral over the sphere
of center in x and with the radius r, so the boundary of Bxr.
Of course, it works the same, but I mean, I thought it was a little more convoluted.
You see, here we didn't use any kind of regular integration formula or stuff like that.
We just straightforwardly take a derivative and then integrate by parts.
In the end, the approaches are equivalent.
I don't know, I just felt this was easier.
The part B, which was optional, consistently showing that if a function is sullarmonic,
then it also satisfies, and continues up to the boundary, of course, and it satisfies
the weak-maximal principle.
The weak, not the strong one.
And this is the point.
Some of you tried to prove the strong one, but here we are not assuming you to be connected.
And on this, I must say one thing, sorry.
But if you think that I did not put an assumption in an exercise, well, it would be much better
that you tell me that either on Wednesday afternoon when there is our online weekly
meeting or by email in any moment, it would be, I mean, I make mistakes, unfortunately.
Everybody does.
So if you think that there are assumptions missing in the exercise, it's much better
that you communicate them to me so that, first of all, you are sure that you understood correctly.
And second, also, your colleagues will, I mean, will get some advantages from knowing
that there is a problem and maybe they will avoid making mistakes.
I don't know.
But usually, if I, I mean, if I write an exercise like this, I'm not assuming you to be connected,
and I want you to prove exactly this statement, not the strong maximal principle.
I never asked you to prove that we cannot attain the maximum inside you unless V is constant.
No.
As I explained before while explaining the solution to the first exercise, here you are
required only to show that the maximum over the closure is equal to the maximum over the
boundary.
This does not imply that V cannot attain the maximum also inside.
It says if V has a maximum point inside you, then its value must be less equal than the
maximum over the boundary, period, that it can be very well equal.
There's no contradiction in that.
So trivially notice that the maximum over the closure is equal to the maximum over the
boundary.
All you have to do is only to prove the reverse inequality.
Okay.
So assume by contradiction that we can find the point inside you such that the value of
V in this x0 is strictly larger than the maximum over the boundary.
If it will be equal, it will be done.
Now by point 2a, we have that the maximum over the boundary, I mean, it's less than
V of x0.
V of x0 will be less equal than the average integral over a ball Bx0r of V of y and dy.
For all, r posits are such that Bx0r is contained in u.
Since u is open, I can always find r positive with this property.
And this, of course, is controlled from above by the maximum over the closure, which is
equal to V of x0.
In other terms, the integral over B of x0r of V of y minus V of x0 in dy should be 0.
Of course, no?
I subtract, and this chain of inequalities becomes a chain of equalities.
But V of y is less equal than V of x0 because it's the maximum.
So this means that this function is a sign, in particular negative, is less equal than
0, but the integral must be 0, so the function must be 0.
So we prove that V of y must be equal to V of x0 for all y in the ball Bx0r.
Now, I found this argument before I had the idea that there was a simpler one, but this
is more by hand, so if you are not so keen in topology or if you don't remember some
statement, it's totally normal, this would be like done by hand.
So the idea is set R, capital R, to be the supremum of R positive such that B of x0r
is inside u.
This is well defined.
We claim that the closure of Bx0r must intersect the boundary of u, must touch it, at least
in a point.
Yes, this would seem reasonable.
By contradiction, let's assume that an intersection is empty.
Okay, but now we know that, of course, the closure of Bx0r is contained in the closure
of u.
No, of course.
No.
It cannot get out of u by the definition of capital R.
Therefore, the closure of Bx0r intersection of the complementary of u must be empty because
it's inside the closure and doesn't touch the boundary.
Now Bx0r closure is compact because it's bounded and closed by assumption.
Rn minus u is closed, of course, because it's complementary of an open set, and therefore
these sets are disjoints and therefore they have positive distance.
Remember, if you have a compact and a closed set, this joint, they must have positive distance.
It's an exercise.
We define D to be this distance, so the infimum of x minus y norm for x in closure of Bx0r
and y in the complementary view.
D must be positive, no?
Because we said that all these things happen, that they don't touch and so on.
Therefore, the closure of Bx0r plus D over 2 intersection Rn minus u should still be
empty.
Yes, because if this is the distance between the closure of Bx0r and the complementary
of u, if I go half this distance, they still don't touch.
Thus, we deduce that Bx0r plus D over 2, not closed, should be inside u.
No?
Because if it doesn't touch the complementary, I take the interior, the interior must be
inside u.
But this contradicts the fact that R is the supremum of the values of small r such that
Bx0r is inside u.
Thus, the closure of Bx0r touches the boundary of u.
But now, v is continuous up to the boundary and v of y is equal to v of x for every y
in Bx0r and for every r from 0 to capital R. This means that v of y must be equal to v
of x0 for all y in the closure of Bx0r, because it is continuous.
And so there exists y0 in Bx0r intersection the boundary such that v of x0 is equal to
v of y0.
And v of y0 is, of course, less equal than the maximum of v over the boundary of u.
And thus, from this chain of inequality, we conclude.
The simpler way to solve the exercise, but which requires a little more topology, is
the following.
So as before, assume you find x0 inside it, v of x0 is the maximum over the closure of
u.
And we call it capital M. We can prove that v of y will be equal to capital M for all
y in Bx0r for all positive such that Bx0 is inside u, again, using the average inequality
on the both.
OK.
We define u m to be the set of point in u, sorry, this is not a closure, it was a mistake,
in u, such that v of x is equal to m.
Then u m is closed and open, closed by definition, because it's a counter-image of a point, and
v is continuous, and open because we proved the error.
So whenever a point x0 belongs to this set, all the balls enter, there exist a small balls
enter at this point, such that all points in this ball belongs to this set.
Wonderful.
U m is not empty, because x0 belongs here by assumption.
Therefore, u m must contain the connected component to which x0 belongs.
This means that v must be constantly equal to m in the connected component of u which
contains x0.
And so by continuity, v is identically equal to m, also in the boundary of this component.
Refresing will move mathematics, symbols, where there exist an open connected set v
containing u, such that v is also containing u m, and so v is equal to m up to the closure
of v.
That's all.
The maximum of v over the boundary of u is less equal than the maximum of v over the
closure, which is equal to the maximum of v over the closure of v, the assumption.
It is equal to the maximum of v over the boundary of v, again, but this is less equal than the
maximum of v over the full boundary of u, because v is a subset of u.
It's not only a subset, a connected component.
And therefore, we are done.
And this is the point, you see.
If u would be connected, I don't have to do all this stretch here.
There would be only one connected component, and so u m is equal to that connected component,
so it's equal to u.
Period.
v is constant.
But in general, u may not be.
Again, the example I told before, take the two segments on the line, 0, 1, 3, 4, u of
x equal to x, and then equal to 1, x in 0, 1, and then 1 in 3, 4.
This is the point.
Now, going on...
No, sorry, it went too fast.
The other two parts seem very quite simple.
So if phi is c2 and is convex, u is c2 and harmonic, prove that phi of u is subharmonic.
Well, the gradient of v is, of course, phi prime against du, then recall the chain rule
for the divergence, the divergence of the product between a scalar function and the
vector valid function is equal to the gradient of g against f plus the g, the divergence
of f.
And so phi prime of u is clearly a scalar.
du is the gradient and the vector valid function.
So the divergence of dv, the Laplacian of v, is equal to phi second of u against du
squared, because I take again the derivative of the gradient of phi prime and gives me
the second derivative of phi times du times du plus phi prime Laplacian of u.
But now Laplacian of u is 0, so it remains this term, but phi second is non-negative
because phi is c2 and convex, and this is non-negative because it's a square of a vector.
So we're done.
The second part was prove that if u is a c3 harmonic function u, then the square of the
norm of the gradient of u is subharmonic.
So clearly v is a u.
dv clearly is equal to twice the Hessian of u against du.
In other terms, dv in the xi is equal to twice the sum over k from 1 to n of du in
the xk, d squared du in the xk dxi.
And so the Laplacian is equal to a sum over j from 1 to n, d in the xi, d in the xiv.
So plug in this definition, the sum over j from 1 to n, twice, who cares, sum over j
from 1 to n, d in the xi, xj, sum over k from 1 to n, du in the xk, d squared du in the
xk dxj.
Taking a derivative here, will give us the double sum over j in k, d squared du dxj dxk,
d squared du dxk dxj, this is du in dxk squared, plus du in dxk times d3u in dxk dxj squared.
This one is the square of the norm of the Hessian, which is defined in the Euclidean
sense.
It should take the square root of the sum of all the components of this matrix, and
of course it's non-negative.
And here instead, by bringing inside the sum in j, we get the sum over k from u in the
xk times d in the xk of the sum in j from 1 to n of d squared du in dxj squared.
And this is the Laplacian.
So we get twice du against the gradient of the Laplacian of u.
But the Laplacian of u is equal to 0, and so the gradient of the Laplacian of u is equal
to 0.
Or in other terms, you can also exchange the operators, if u is harmonic, then also its
derivatives are harmonics.
And so, we are done.