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