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...


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.

  • $\begingroup$ 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
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    $\begingroup$ @sgtswordfish. I need to check my copy when I get home later today. $\endgroup$ – md2perpe May 11 '20 at 11:48
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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