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

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?

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 $f$.

• Can you provide a reference to where the Lp-norm of a distribution is defined? I am having a hard time finding one. Oct 29, 2016 at 17:37
• Hmm, don't know of a reference - duality is just the best setup for $L^p$ norms…
– Dirk
Oct 29, 2016 at 17:41
• To clarify, you are saying that since distributions are continuous linear functionals, they have the dual norm ${\displaystyle \left\|f\right\|=\sup\{\left|f(x)\right|:x\in X,\left\|x\right\|\leq 1\}}$? Oct 29, 2016 at 17:48
• If this norm is finite, $T$ is a continuous linear functional on $L^q$, i.e. it can be identified with an element in the dual space of $L^q$.
– Dirk
Jun 19, 2019 at 15:51
• $\mathcal{D}$ are the test functions. The distributions are $\mathcal{D}'$.
– Dirk
Nov 12, 2021 at 6:23

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.

• You mean to say that $L^p(\mathbb R^n)$ is the collection of distributions satisfying $T(g) = \int_{\mathbb R^n}fgdx$ and such that $||f||_p < \infty$? May 12, 2013 at 18:07
• Yes, that's it. May 12, 2013 at 18:08
• Thanks for your answer, which I greatly appreciate. However, I chose to accept Dirk's answer since it does not require an a-priori restriction on the form of $T$. May 12, 2013 at 23:31
• There is no problem @ArthurTrask May 13, 2013 at 0:12
• What do you mean by writing "$\int_\mathbb{R} f(x)dx=\infty$"? If f "is" the Dirac Distribution $\delta$ the "integral" of the dirac is 1. If it is just a function defined as $\mathbb{R}\setminus\{0\}$, $f(0)=\infty$ the classical Lebesgue integral would be not defined or defined as zero: en.wikipedia.org/wiki/Lebesgue_integration#Definition Nov 11, 2021 at 18:28