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How to prove that functions in $L^p(\mathbb{R}^n),1 \leq p \leq \infty$, are tempered distributions.

J.R.
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3 Answers3

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The above answers have shown that if $f \in L^p$ and $\varphi \in \mathcal{S}$, then $\int f \varphi$ is well defined and finite.

To show that it is a tempered distribution (that is, the map $\varphi \mapsto \int f \varphi$ is continuous on $\mathcal{S}$), we need to show that there is a constant $C > 0$ and $M, L \in \mathbb{N}_0$ such that for all $\varphi \in \mathcal{S}(\mathbb{R}^n)$, $$ |\int f\varphi| \leq C \sum_{k=0}^M \sum_{|\alpha| \leq L} \|D^\alpha (1 + |x|)^k \varphi(x)\|_\infty$$

Now, $$ \left|\int f \varphi\right| \leq \int |f(x)|(1 + |x|)^{N}(1 + |x|)^{-N}|\varphi(x)|\,dx $$ $$\leq \|(1 + |x|)^N\varphi(x)\|_\infty \int |f(x)|(1 + |x|)^{-N}dx $$ $$\leq \|(1 + |x|)^N\varphi(x)\|_\infty \|f\|_{L^p} \|(1 + |x|)^{-N}\|_{L^q}$$

Choose $N$ large enough so that $\|(1 + |x|)^{-N}\|_{L^q} < \infty$ and now the result holds with $L = 0, M = 0$ and $C = \|f\|_{L^p}\|(1 + |x|)^{-N}\|_{L^q}$.

Anu
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Let $f\in L^p(\mathbb{R}^n)$. The canonical distribution corresponding to $f$ is defined by $d_f:\mathcal{S}(\mathbb{R}^n)\rightarrow \mathbb{C}$, $g\mapsto \int_{\mathbb{R}^n} fg$.

Now check that $d_f$ is an element of $\mathcal{S}^\prime(\mathbb{R}^n)$. It is clear that it is a linear map. So it remains to check continuity. Namely, we have to check that

$$\|d_f\|_{\mathcal{S}^\prime(\mathbb{R}^n)}=\sup_{\|g\|_{\mathcal{S}}=1} \left|\int_{\mathbb{R}^n} fg\right|<\infty$$

But let $g$ be a Schwartz function of Schwartz norm $1$. Then it is in all $L^p$ spaces, in particular in $L^q$ with $1/p+1/q=1$, so the claim follows by Hölder's inequality:

$$\int_{\mathbb{R}^n} |fg| \le \|f\|_p \|g\|_q<\infty$$

J.R.
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If $f\in L^p(\mathbb R^n)$ and $\varphi\in\mathscr S(\mathbb R^n)$, then $$ \ell(\varphi)=\int_{\mathbb R^n}f\,\varphi, $$ is definable, since $f\varphi\in L^1(\mathbb R^n)$, as $\varphi\in L^q(\mathbb R^n)$, for all $p$, and $$ \lvert \ell(\varphi)\rvert\le\|\varphi\|_{p'}\|\,f\|_p, $$ where $\frac{1}{p}+\frac{1}{p'}=1$.

All the other properties are straight-forward.