# If a Radon measure is a tempered distribution, does it integrate all Schwartz functions?

The question might at first sight sound like the answer is trivially "yes", so let me clarify the question a bit. Consider given a nonnegative Radon measure $\mu$ on $\mathbb{R}^n$. Let $\mathcal{D}(\mathbb{R}^n)$ denote the space of real test functions, that is, the space of infinitely differentiable functions with compact support from $\mathbb{R}^n$ to $\mathbb{R}$. We may then define the operator $T:\mathcal{D}(\mathbb{R}^n)\to\mathbb{R}$ by $$T(\varphi) = \int_{\mathbb{R}^n} \varphi(x) d\mu(x).$$ This is then a distribution, in the sense of being a continuous linear functional $\mathcal{D}(\mathbb{R}^n)\to\mathbb{R}$, when $\mathcal{D}(\mathbb{R}^n)$ is given its usual topology based on a family of norms as in Rudin's "Functional analysis".

Now let $\mathcal{S}(\mathbb{R}^n)$ denote the space of real Schwartz functions, meaning the space of rapidly decreasing functions, endowed with its usual topology, again see Rudin's book for details.

$\textbf{Assume the following:}$ That $T$ can be extended to a linear functional on $\mathcal{S}(\mathbb{R}^n)$ which is continuous in the topology of $\mathcal{S}(\mathbb{R}^n)$.

$\textbf{My question is this:}$ Does it holds that $$T(\varphi) = \int_{\mathbb{R}^n} \varphi(x) d\mu(x).$$ for all $\varphi\in\mathcal{S}(\mathbb{R}^n)$?

$\textbf{Some remarks:}$ The problem is that while we know that $T$ extends from $\mathcal{D}(\mathbb{R}^n)$ to $\mathcal{S}(\mathbb{R}^n)$, we do not know that the integral form of the operator carries over from $\mathcal{D}(\mathbb{R}^n)$ to $\mathcal{S}(\mathbb{R}^n)$. In fact, we do not even know that all functions in $\mathcal{S}(\mathbb{R}^n)$ are integrable with respect to $\mu$.

A natural first approach to the problem would be to consider some kind of approximation argument. For example, take $\varphi\in\mathcal{S}(\mathbb{R}^n)$. Assume that $\varphi\ge0$. Using convolutions, we may then construct a sequence $(\varphi_n)$ in $\mathcal{D}(\mathbb{R}^n)$ which converges monotonely and in $\mathcal{S}(\mathbb{R}^n)$ to $\varphi$. This then yields

$$\int_{\mathbb{R}^n} \varphi(x) d\mu(x) = \lim_n \int_{\mathbb{R}^n} \varphi_n(x) d\mu(x) = \lim_n T(\varphi_n),$$ where the limit a priori may be infinite. However, as we have assumed that $T$ extends to $\mathcal{S}(\mathbb{R}^n)$, the limit of $T(\varphi_n)$ is finite, and so $\varphi$ is integrable with respect to $\mu$. This should show that the integral form of the operator is preserved for all nonnegative Schwartz functions.

For a general Schwartz function $\varphi$, the natural thing would be to write $\varphi = \varphi^+-\varphi^-$, where $\varphi^+$ and $\varphi^-$ are the positive and negative parts of $\varphi$, respectively. However, while these two functions are nonnegative, they are not Schwartz functions, nor test functions for that matter, and so we cannot apply $T$ to them.

• Isn't $e^x dx$ a Radon measure? And you cannot integrate every tempered function against $e^x$. – Vobo Oct 30 '13 at 7:00
• @Vobo: The assumption is that the associated distribution is tempered which is not the case for the measure you propose. – Jochen Oct 30 '13 at 8:46
• Yes, I should read the title properly... sorry. – Vobo Oct 30 '13 at 14:53
• As far as I know, a Radon measure is a distribution of order 0 and extends continuously to the space $C_c(\mathbb{R})$ still with the valid formula $T(\varphi)=\int \varphi d\mu$. And for just continuous test functions $\varphi$, $\varphi^+$ and $\varphi^-$ are continuous test functions too, so that you can repeat your previous arguments as desired. – Vobo Oct 30 '13 at 22:00
• @AlexanderSokol Don't you think the argument in my previous comment answers the problem? – Vobo Nov 5 '13 at 6:58

Suppose we can find two non-negative Schwartz functions $f,g$ such that $\phi = f-g$, then we would be done. Hence it is enough to prove that
Lemma: Given any Schwartz function $f:\mathbb{R}^n\to \mathbb{R}$, there is a Schwartz function $g:\mathbb{R}^n \to \mathbb{R}$ with $g\geq 0$ and $|f(x)|\leq g(x)$ for all $x\in \mathbb{R}^n$.
Proof: Define $g_1(x) = \sup \{\lvert f(y) \rvert \; | \; \lvert y-x \rvert \leq 1 \}$. Let $\phi \geq 0$ be a bump function supported on $B(0,1)$ with $\int \phi = 1$. Define $g = g_1 * \phi$. Clearly $g(x)\geq 0$ and $g(x)\geq |f(x)|$ for all $x\in \mathbb{R}^n$ and it is easy to see that $g$ is Schwartz.