WEBVTT - autoGenerated
00:00:00.000 --> 00:00:08.000
So, we pass now to the main consequence of Young's inequality, that is Helder's inequality
00:00:08.000 --> 00:00:15.000
for the LP norm. It states that if P and Q are conjugate exponents, which means 1 plus
00:00:15.000 --> 00:00:22.000
Q plus 1 over P is equal to 1, and in this case we can take P and Q even to be 1 and
00:00:22.000 --> 00:00:29.000
infinity respectively, then if we have F in LP and G in LQ, their product will be in L1
00:00:29.000 --> 00:00:34.000
and their L1 norm of the product is controlled by the product between the LP norm of F and
00:00:34.000 --> 00:00:41.000
the LP norm, the Q norm of G. The case P equal 1 and Q infinity and vice versa is trivial
00:00:41.000 --> 00:00:46.000
because simply the absolute value of G of X is controlled from above by the L infinity
00:00:46.000 --> 00:00:51.000
norm of G, almost everywhere, but negligible sets don't count when integrating with respect
00:00:51.000 --> 00:00:56.000
to the Lebesgue measure, so we bring it outside and this is the L1 norm of F, so we are done.
00:00:56.000 --> 00:01:01.000
More interestingly, this is the case of P and Q inside one infinity without counting
00:01:01.000 --> 00:01:07.000
the extremes. Without loss of generality, you can assume that neither of the LP norm
00:01:07.000 --> 00:01:13.000
of F and the Q norm of G are zero, otherwise everything is trivial because F and G would
00:01:13.000 --> 00:01:19.000
be zero almost everywhere, so fix X in U, apply Young's inequality by taking as A, the
00:01:19.000 --> 00:01:24.000
absolute value of F of X divided by the LP norm of F and B, the absolute value of G of
00:01:24.000 --> 00:01:30.000
X divided by the IQ norm of F, so that we get this inequality, and now we simply integrate
00:01:30.000 --> 00:01:39.000
over U, both sides. On the left-hand side, sorry, we get basically the L1 norm of F times
00:01:39.000 --> 00:01:47.000
G divided by the LP norm of F times the IQ norm of G, and on the right-hand side, well,
00:01:47.000 --> 00:01:52.000
when you take the integral here, you get the LP norm of F to a power of P divided by the
00:01:52.000 --> 00:02:00.000
LP norm of F to a power of P, so it simplifies, remains one over P. Same argument here, you
00:02:00.000 --> 00:02:07.000
get one over Q, and this is one by assumption. By the way, this is why you need to put this
00:02:07.000 --> 00:02:13.000
condition, or by Young's inequality, of course, and so we are done. So, Young's inequality
00:02:13.000 --> 00:02:19.000
is very relevant because there's many generalization and consequences. Let's start. First of all,
00:02:19.000 --> 00:02:25.000
it's not necessary that on the right-hand side here, there is one. You can take another
00:02:25.000 --> 00:02:33.000
exponent P, sorry, and have one over P. So, if P1, P2, and P are of this relation, and
00:02:33.000 --> 00:02:39.000
F1 is in LP1, and F2 is in LP2, then the product will be in LP, where the LP norm of the product
00:02:39.000 --> 00:02:47.000
is controlled by the product of the LP1 and LP2 norm, respectively. Again, assume at first
00:02:47.000 --> 00:02:55.000
that neither of these exponents is plus infinity. Then, P1 over P and P2 over P are conjugate
00:02:55.000 --> 00:03:00.000
exponents. It comes immediately. Therefore, you can apply Hölder's inequality to show
00:03:00.000 --> 00:03:08.000
that the integral over U of the product between F1 to the P1, so far, and F2 to the P2 is controlled
00:03:08.000 --> 00:03:16.000
by what you should expect on the right-hand side. So, we are done. If instead, is it possible
00:03:16.000 --> 00:03:24.000
that one or many of these is plus infinity? Well, if P2 is infinity, then clearly P1 must
00:03:24.000 --> 00:03:30.000
be equal to P, and vice versa. If P1 is infinity, then P2 must be equal to P. And inequality
00:03:30.000 --> 00:03:35.000
is trivial in this case, because you're basically saying that F2 is bound, that is essentially
00:03:35.000 --> 00:03:40.000
bounded, and so you bring out the LP norm, as you did before in the easy case of the
00:03:40.000 --> 00:03:47.000
Hölder's inequality. Notice that if instead is a P to be plus infinity, here you get 0,
00:03:47.000 --> 00:03:51.000
and so the only possible case is that both P1 and P2 are infinity. And so the inequality
00:03:51.000 --> 00:03:56.000
becomes completely trivial, because it becomes the LP norm of F1 as true is controlled by
00:03:56.000 --> 00:04:06.000
the LP norm of F1 times the LP norm of F2. It's very easy to check. Very well. Then,
00:04:06.000 --> 00:04:12.000
you can generalize the Hölder's inequality to many exponents, any number of exponents,
00:04:12.000 --> 00:04:17.000
provided that they are conjugate in the sense that the sum over 1 over pi j, for j from
00:04:17.000 --> 00:04:27.000
1 to k, is equal to 1 over P. Then, if we take k functions, Fj, each one in Lpj, then
00:04:27.000 --> 00:04:33.000
their product will belong to LP, and the LP norm of a product will be less equal than
00:04:33.000 --> 00:04:41.000
the product of the Lpj norm of Fj for j from 1 to k. Again, this is proved arguing as before
00:04:41.000 --> 00:04:46.000
by induction in this case. The k is k equal to 2 is Hölder's inequality, so it has been
00:04:46.000 --> 00:04:53.000
done already. We proved the general case by assuming that the generalized Hölder's inequality
00:04:53.000 --> 00:05:01.000
holds for k minus 1. And then, we noticed that 1 over P minus 1 over Pk is simply the
00:05:01.000 --> 00:05:07.000
sum of 1 over Pj for j from 1 to k minus 1. Clearly, no? We are taking away one term.
00:05:07.000 --> 00:05:14.000
But this can be also written as Pcspk minus P over Ppk. And then, the idea is we apply
00:05:14.000 --> 00:05:18.000
Hölder's inequality to the function of the product, which are given by the product from
00:05:18.000 --> 00:05:27.000
j from 1 to k minus 1 over Fj and Fk, so the remaining one. And the idea is, for Fk, we
00:05:27.000 --> 00:05:34.000
choose the exponent Pk over P, and for the product, what remains. So now, we have the
00:05:34.000 --> 00:05:38.000
integral over u of the product from j from 1 to k of the absolute value of j to the power
00:05:38.000 --> 00:05:43.000
P. Applying Hölder's inequality will give us integral over u of absolute value of k
00:05:43.000 --> 00:05:54.000
to the power Pk, to all 2 P over Pk. And this is good, because this is the LPk norm of Fk.
00:05:54.000 --> 00:05:59.000
And here, instead, we have the product to this exponent, you see. Here, this power P.
00:05:59.000 --> 00:06:07.000
So, Ppk over Pk minus P. But this exponent is clearly the inverse of this. So, here we
00:06:07.000 --> 00:06:13.000
apply in the induction hypothesis. So, the generalized Sinequity for k minus 1 functions
00:06:13.000 --> 00:06:19.000
and exponents. And we are done. Another interesting result is the inclusion between LP spaces
00:06:19.000 --> 00:06:25.000
in the case the domain u as a finite measure. In this case, we have that Lq is inside LP
00:06:25.000 --> 00:06:31.000
with continuous inclusion. For all q less equal than P, between 1 and infinity. Actually,
00:06:31.000 --> 00:06:38.000
PRP could be taken only large and non-negative, larger than CL. And the inclusion is given
00:06:38.000 --> 00:06:44.000
by this inequality. The continuity of the inclusion is given by this inequality. The
00:06:44.000 --> 00:06:49.000
LP norm of F is controlled by the measure of u to a power 1 over P minus 1 over q times
00:06:49.000 --> 00:06:56.000
the iq norm of F for all F in Lq. The trick is apply again here the Sinequity with exponents
00:06:56.000 --> 00:07:01.000
q over P and q over q minus P. Notice that since q is larger equal than P, this is larger
00:07:01.000 --> 00:07:10.000
equal than 1, and so things make sense. Indeed, if we take F in Lq, we have that the integral
00:07:10.000 --> 00:07:14.000
over u of F to the power of P will be controlled by the integral over u of F to the power of
00:07:14.000 --> 00:07:20.000
q. The reason why it remains is basically I'm saying, sorry, here I'm seeing F to the power
00:07:20.000 --> 00:07:26.000
of P times 1. And so I'm raising the absolute value of F to the power of P to the power
00:07:26.000 --> 00:07:33.000
q over P in order to get absolute value of F to the power of q times integral of 1, so
00:07:33.000 --> 00:07:40.000
measure of u to the power of 1 minus P over q. Yeah, sure. To be completely fair here,
00:07:40.000 --> 00:07:46.000
we didn't consider the case of q infinity, but if, again, if q is equal to infinity,
00:07:46.000 --> 00:07:52.000
everything is trivial. We don't really need to do the explicit calculation of this inequality.
00:07:52.000 --> 00:08:10.000
We just take out the... Sorry. If q is infinity, you simply take out the supremum of F and it
00:08:10.000 --> 00:08:17.000
remains the measure of u to the power of 1 minus P. Also, this can be seen as a consequence
00:08:17.000 --> 00:08:24.000
of yes and inequality once we use the convex map F of t equal to t to the q over P. Another
00:08:24.000 --> 00:08:30.000
incredibly important consequence of Hölder's inequality is Minkowski's inequality, which is
00:08:30.000 --> 00:08:35.000
basically the triangle inequality for the LP semi norm or norm whenever you make the
00:08:37.000 --> 00:08:44.000
cosintation. If P is in one infinity and F, g are in LP, then the LP norm of F plus g is
00:08:44.000 --> 00:08:51.000
less equal than the LP norm of F plus the LP norm of g. As a consequence, immediately, this
00:08:52.000 --> 00:08:58.000
norm, this is actually semi norm on LP. Again, remember, it says only semi norm because we
00:08:58.000 --> 00:09:05.000
didn't use the cosintation with respect to the equivalence relation, which says that it identifies
00:09:05.000 --> 00:09:11.000
two functions which are equal almost everywhere. The proof is trivial in the sense that clearly
00:09:11.000 --> 00:09:19.000
this norm, this LP content is one homogeneous and this is the triangle inequality.
00:09:20.000 --> 00:09:27.000
This was given as an exercise. Actually, we can give a generalized statement because you see
00:09:27.000 --> 00:09:33.000
here clearly F, it works for the sum of F and g, but it can work for any finite sum.
00:09:34.000 --> 00:09:41.000
Actually, it works also for a continuous sum. More precisely, if we take a function F from
00:09:41.000 --> 00:09:49.000
the tensor product, the cross product of two open sets, U and V, in possibly in different
00:09:49.000 --> 00:09:56.000
dimensional spaces, and we ask it to be measurable and such that for almost every fixed index in U,
00:09:56.000 --> 00:10:07.000
Fx is a function of y is in one of V, then for every P, one in feet, the LP norm of U
00:10:07.000 --> 00:10:15.000
of the integral of V of F dot y in dy is controlled by the integral of V of the LP norm of U of F dot
00:10:15.000 --> 00:10:23.000
y. Again, you are exchanging this continuous version of the sum and the LP norm. In particular,
00:10:23.000 --> 00:10:28.000
if F is non-negative and we assume that this quantity is finite, which means F should be L1
00:10:28.000 --> 00:10:38.000
in V and LP in U, then it's L1 norm in V, which is this integral here, will belong to Lp of omega.
00:10:40.000 --> 00:10:46.000
And also this was given as an exercise. Finally, coming to another important result is the
00:10:46.000 --> 00:10:53.000
interpolation of LP spaces. If we take P less equal than Q between 1 and infinity and F in
00:10:53.000 --> 00:11:02.000
LP intersectional Q, then actually F belongs to LR for all R between P and Q with the LR norm of F
00:11:02.000 --> 00:11:07.000
is controlled by the LP norm of F to the power theta times the LQ norm of F to the power 1 minus theta,
00:11:07.000 --> 00:11:13.000
where theta is 0, 1. So this is 1 over R equal to theta over P plus 1 minus theta over Q.
00:11:14.000 --> 00:11:20.000
This is a convex combination of exponents. So again, if Q is infinity, the proof is simpler
00:11:20.000 --> 00:11:27.000
because if Q is infinity here, this term vanishes and theta must be P over R. So the integral of
00:11:27.000 --> 00:11:35.000
U of F to the power R absolute value will be decomposed as integral over U of absolute value
00:11:35.000 --> 00:11:40.000
of U of F to the power R theta times f lose over U of F to the power R times 1 minus theta.
00:11:42.000 --> 00:11:50.000
R theta is P, the assumption, so this will give us the LP norm to the power P, which is R theta,
00:11:50.000 --> 00:11:56.000
once we bring out the L infinity norm of this term because it's bound, F is bound.
00:11:57.000 --> 00:12:01.000
And so well done. If instead the Q is less than infinity, again notice that
00:12:01.000 --> 00:12:07.000
thanks to this relation, by dividing, we have that P over R theta and Q over R times 1 minus theta
00:12:07.000 --> 00:12:12.000
are conjugate exponents. So again, we can apply U holder's inequality by using the same trick
00:12:12.000 --> 00:12:24.000
as above and we are done again. Now let us come to a somehow very deep result concerning some weak
00:12:25.000 --> 00:12:32.000
continuity properties of L1 log function, to put it in naive terms, which is Lebesgue
00:12:32.000 --> 00:12:41.000
differentiation theorem. If F is in L1 log for almost every x, x0, sorry, in Rn, so outside of
00:12:41.000 --> 00:12:48.000
a Lebesgue negligible set, we have that the limit as R goes to 0 of the average integral of a busy
00:12:48.000 --> 00:12:57.000
xR, the x0 R of the absolute value of f of x minus f x0 is 0. And therefore, the limit of this
00:12:57.000 --> 00:13:04.000
average of the average of the ball centered in x0 with various R of f of x is actually f of x0.
00:13:05.000 --> 00:13:12.000
In addition, if f is LP log for some P larger than 1 less than infinity, then we have the limit
00:13:12.000 --> 00:13:21.000
4 also for the power P, as one might find believable. A point x0 for which 4 is satisfied
00:13:21.000 --> 00:13:28.000
is called as a back point or f.