In this lecture we define the Young‘s integral when and with . The cornerstone is the following Young-Loeve estimate.
Theorem: Let and . Consider now with . The following estimate holds: for ,
Proof: For , let us define
We have for ,
As a consequence, we get
Let now . We claim that is a control. The continuity and the vanishing on the diagonal are obvious to check, so we just need to justify the superadditivity. Let , we have from Holder’s inequality,
We have then
For , consider then the control
If is such that , we can find a such that , . Indeed, the continuity of forces the existence of a such that . We obtain therefore
which implies by maximization,
By iterating times this inequality, we obtain
It is now clear that:
Sending , finishes the proof
It is remarkable that the Young-Loeve estimate only involves and . As a consequence, we obtain the following result whose proof is let to the reader:
Proposition: Let and with . Let us assume that there exists a sequence such that in and a sequence such that in , then for every , converges to a limit that we call the Young’s integral of against on the interval and denote .
The integral does not depend of the sequences and and the following estimate holds: for ,
The closure of in is and we know that . It is therefore obvious to extend the Young’s integral for every and with and the Young-Loeve estimate still holds
From this estimate, we easily see that for and with the sequence of Riemann sums
will converge to when the mesh of the subdivision goes to 0. We record for later use the following estimate on the Young’s integral, which is also an easy consequence of the Young-Loeve estimate (see Theorem 6.8 in the book for further details).
Proposition: Let and with . The integral path is continuous with a finite -variation and we have