# Tempered distributions : why $\int_{\mathbb R^n}\varphi(x)dx =\langle1,\varphi\rangle$?

Let $$X$$ be a Banach space.

For $$f\in \mathcal S'(\mathbb R^n,X)$$, we define Fourier transform of $$f$$ as $$\langle \mathcal Ff,\varphi\rangle:=\langle f,\mathcal F\varphi\rangle$$ for each $$\varphi\in\mathcal S(\mathbb R^n).$$

And define dirac delta $$\delta$$ as $$\langle \delta,\varphi\rangle:=\varphi(0)$$ for each $$\varphi\in \mathcal S(\mathbb R^n).$$

Then, show that $$\mathcal F\delta =1.$$

Proof

For each $$\varphi\in\mathcal S(\mathbb R^n),$$ $$\langle \mathcal F\delta,\varphi\rangle =\langle\delta,\mathcal F\varphi\rangle =\mathcal F\varphi(0) =\int_{\mathbb R^n}\varphi(x)dx =\langle1,\varphi\rangle.$$

Thus $$\mathcal F\delta=1.$$

I don't understand the last part : $$\int_{\mathbb R^n}\varphi(x)dx =\langle1,\varphi\rangle.$$

I read some books about distribution but those book says $$\int_{\mathbb R^n}\varphi(x)dx =\langle1,\varphi\rangle$$ without explanation.

Why does this hold ? Is this the definition of $$1$$ of tempered distributions ?

• It holds by definition. Jun 9, 2023 at 16:51

By definition, a distribution, $$f$$, acts on the space of test functions via integration. That is, $$\langle f,\varphi\rangle:=\int_{\mathbb{R}^n} f\varphi$$ for a test function $$\varphi$$. So they’re just noticing that $$\int \varphi=\int 1\cdot \varphi:=\langle 1,\varphi\rangle$$ I.e. integration is just integration against the constant function one which is by definition the action of the distribution defined by the constant one function on the space of test functions.
• Not all distributions act by integration. In this case it works because to every locally integrable function a distribution that acts by integration can be associated. Of course this is the case here for the constant $1$ function.