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For $p$ = 1, it's clear because for $f \in L^1(\mathbb R^n)$, one can define a tempered distribution by $T_{f}(\varphi) := \int_{\mathbb R^n} f(x) \varphi(x) dx$.

Apparently, one can make the argument that $\mathcal S(\mathbb R^n) \subset L^p(\mathbb R^n)$ implies that the for the dual spaces the opposite inclusion holds. Why is that?

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  • $\begingroup$ Re: the first point, why is it any different for $p>1$? $\endgroup$ Mar 27 '18 at 20:45
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    $\begingroup$ An intuitive answer is as follows: any bounded linear functional on $L^p$ can be restricted to $\mathcal{S}$ and the worst that can happen is that the norm decreases. Thus any element of $(L^p)'$ can be restricted to $\mathcal{S}$, and is therefore an element of $\mathcal{S}'$. There are different norms running around, so some care needs to be taken, but this is the basic intuition that you maybe should have. $\endgroup$
    – Xander Henderson
    Mar 27 '18 at 20:47
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Regarding the "abstract nonsense" statement "$A \subset B \Rightarrow B' \subset A'$", the important thing there is that the inclusion mapping $i : A \to B,i(x)=x$ is continuous. Given $F \in B'$, the restriction of $F$ to $A$ is $F \circ i$. If $i$ is continuous, then $F \circ i$ will be continuous. So the restriction is in $A'$. We identify the restriction with the original functional $F$.

This identification is an abuse of notation: strictly speaking an element of $B'$ can never be an element of $A'$ or vice versa. But it is a reasonable abuse of notation, especially in cases like this one, in which $A$ is dense in $B$, so that the restriction uniquely specifies $F$. This situation in which $A$ is a dense subset of $B$ with a continuous inclusion mapping is very common in functional analysis.

You can expand this out into a statement about norms and seminorms in this particular case, but this is really the main point.

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  • $\begingroup$ Just a little quibble: in general, the dual of an inclusion need only by a quotient map. To have the dual be injective requires dense image (in the category of locally-convex TVSs, so that duals always separate points, by Hahn-Banach). $\endgroup$ Mar 28 '18 at 0:57
  • $\begingroup$ @paulgarrett I think what you're saying is, identifying $F$ with $F \circ i$ is not just an abuse of notation but is just not valid if $A$ is not dense. And in that regard I agree. Although the preceding still goes through: $F \mapsto F \circ i$ is still a continuous functional on $A'$ if $i$ is just continuous. $\endgroup$
    – Ian
    Mar 28 '18 at 1:08
  • $\begingroup$ Oh, yes, indeed, the dual map is still continuous... I didn't mean to seem to cast any doubt on that...! $\endgroup$ Mar 28 '18 at 2:03
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\begin{align*} \left|\int f(x)\varphi(x)dx\right|&=\left|\int\dfrac{f(x)}{(1+|x|^{n})}(1+|x|^{n})\varphi(x)dx\right|\\ &\leq\left(\sup_{x\in{\bf{R}}^{n}}(1+|x|^{n})|\varphi(x)|\right)\int\dfrac{|f(x)|}{(1+|x|^{n})}dx\\ &\leq\left(\sup_{x\in{\bf{R}}^{n}}(1+|x|^{n})|\varphi(x)|\right)\|f\|_{L^{p}}\left\|\dfrac{1}{(1+|\cdot|^{n})}\right\|_{L^{q}({\bf{R}}^{n})}. \end{align*}

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