How do we define the $L^p$ norm of a tempered distribution?

Question

I am finishing up a class on function theory and I am trying to reconcile a few statements that I have seen.

Let us define $L^p(\mathbb R^n)$ to be the set of measurable functions $f$ so that $\int_{\mathbb R^n} |f|^p dx < \infty$, with the norm $||f||_p = \Big(\int_{\mathbb R^n } |f|^p dx\Big)^{1/p}$.

I have also seen it written that $L^p(\mathbb R^n)$ is the set of all tempered distributions $f \in S'$ satisfying $ ||f||_p < \infty$. Here $S$ is the collection of rapidly decreasing complex valued functions on $\mathbb R^n$, and $S'$ is the space of continuous linear functionals on $S$ (tempered distributions).

I would like to show that these two definitions "are the same". However, I am at a disadvantage, since I don't even see how to define the $||\cdot||_p$ norm on the space of tempered distributions.

Thus, how do we define the $L_p$ norm of a tempered distribution?

Answer 1

Distributions act on the space $\mathcal{D}$ of test functions (infinitely differentiable with compact support). For all $p$ it holds that $\mathcal{D}\subset L^p$ and hence, it makes sense to take the $p$-norm test functions for any $p$. Now take some $p$ and its conjugate $q$ (i.e. $p+q=pq$). For any distribution $T$ we can define $$ \|T\|_p = \sup\{ T(\phi)\ :\ \|\phi\|_q\leq 1\} $$ which may or may not be finite.

Note that this produces the $p$ norm of a function $f$ when applied to the distribution induced by $f$.

Answer 2

You cannot define $\|\cdot\|_p$ in all space of tempered distributions. In general, if $T$ is a distribution, for the value $\|T\|_p$ make sense, we need to identify $T$ with a function defined in $\mathbb{R}^n$. There are some distributions that can not be identified with functions. On the other hand, the distributions that can be identified with functions and satisfies $\|T\|_p<\infty$, constitute the space $L^p$.

Consider for example the Dirac Distribution $\delta$. If you were to identify $\delta$ with some function $f$, then $f(x)=0$ in $\mathbb{R}\setminus\{0\}$, $f(0)=\infty$ and $\int_\mathbb{R} f(x)dx=\infty$. So, you can see that the notion of distribution, generalizes the notion of function.