1

I'm trying to understand the proof of the following theorem in Evans's PDE book.

enter image description here enter image description here

The first step is to rewrite the bilinear form using the Riesz representation theorem: enter image description here

Then the properties of the operator $A$ are proved: enter image description here enter image description here enter image description here

Eventually the Riesz reprensentation theorem is used again to give the existence: enter image description here

The very last step gives the uniqueness: enter image description here

Could anyone identify where the one-to-one property of $A$ is used? I suspect that that is redundant in the proof.

  • The property could have been used to prove uniqueness of the vector, but it appears the argument was repeated to show uniqueness directly, at a later point. – Disintegrating By Parts Dec 24 '16 at 12:15
  • @TrialAndError: This is what this question asked and I think step 5 only gives the existence not the uniqueness. (If one applies the one-to-one property of $A$ in step 5 though, one should get the uniqueness only for those solutions obtained via the Riesz representation of $f$.) –  Dec 24 '16 at 13:03
  • The 1-1 property of A could be used to establish uniqueness. But, as I wrote in the first comment, it appears they repeated the argument, one for uniqueness, and once for 1-1. – Disintegrating By Parts Dec 24 '16 at 14:25
  • Au=f. Because $A$ is 1-1, then $u$ is unique. – Disintegrating By Parts Dec 24 '16 at 14:42
  • I don't see how you can conclude that $Au=f$. –  Dec 24 '16 at 14:44
  • I'm assuming $H$ is a Hilbert space. $f$ is identified with a vector. – Disintegrating By Parts Dec 24 '16 at 14:46
  • @TrialAndError: Fair enough. $B[u,v]=\langle f,v\rangle$ if and only if $(Au,v)=(w,v)$. Now the 1-1 property of $A$ implies that such $u$ must be unique. I agree with you now. Thanks! –  Dec 24 '16 at 14:53

1 Answers1

0

It is indeed redundant as @TrialAndError pointed out in the comment. The 1-1 property of $A$ can be used in Step 5 to show the uniqueness given by Step 6. In the whole proof, one can either drop the 1-1 property of $A$ or Step 6. Evans might just think it is clearer for students to put in the extra step.