0
$\begingroup$

Here is how folland introduces the construction of continuous operators on distributions:

p284 of folland

I understand that the adjoint/equality thing is needed so that the constructed operator is an extension of the original operator.

However, I'm confused about the "continuity of $T'$ guaranteeing continuity of $T$." My question:

For continuity of $T$, why do we need continuity or linearity of $T'$?

To me, it seems that continuity of $T$ only requires that the adjoint $T'$ be a (set) map of the space of test functions $C_c^\infty$ into itself, since if $F_j \to F$ as distributions (ie pointwise on test functions, because weak* topology)

$$ \langle TF_j, \phi \rangle := \langle F_j, T' \phi \rangle \to \langle F, T' \phi \rangle =: \langle TF, \phi \rangle. $$

Actually, based on the above, it seems you could just define a continuous operator $T$ on $D'$ if you have such a set map $T'$.

What am I missing here? I haven't actually checked it, but I bet linearity of $T$ requires linearity of $T'$ was linear...but for continuity...

$\endgroup$
1
$\begingroup$

In which book by Folland did you read this?

I agree with you that continuity of $T'$ seems not be needed for continuity of $T$ extended to $\mathcal{D}'.$ However, I think that it is needed for $TF$ to be a distribution, i.e. for $TF$ to be continuous:

Let $\varphi_j \to \varphi$ in $C_c^\infty$ and assume that $T'$ is continuous. Then, $ \langle TF, \varphi_j \rangle = \langle F, T'\varphi_j \rangle \to \langle F, T'\varphi \rangle = \langle TF, \varphi \rangle ,$ i.e. $TF$ is continuous.

$\endgroup$
7
  • $\begingroup$ ahh you're totally right! and this is page 284 of folland's "real analysis: modern techniques and their applications" $\endgroup$ – sweetpotato May 11 '20 at 11:28
  • 1
    $\begingroup$ @sgtswordfish. I need to check my copy when I get home later today. $\endgroup$ – md2perpe May 11 '20 at 11:48
  • 1
    $\begingroup$ I cannot find that comment in my copy of the book, which must be the first edition. (Image and PDF on Scribd) $\endgroup$ – md2perpe May 11 '20 at 19:25
  • 1
    $\begingroup$ I sent an email to Prof. Folland asking him if we miss something, if we misunderstand what he writes, or if this is an actual error in the book. $\endgroup$ – md2perpe May 11 '20 at 21:35
  • 1
    $\begingroup$ Gerald Folland has now answered my question: "You're right: the place where the continuity of T' is really needed is in guaranteeing that the formula <TF, phi> = <F, T'phi> defines TF as a distribution." $\endgroup$ – md2perpe May 12 '20 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.