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 $\mathcal F\delta =1.$


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 ?

  • 1
    $\begingroup$ It holds by definition. $\endgroup$
    – William M.
    Jun 9, 2023 at 16:51

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – jd27
    Jun 9, 2023 at 17:14
  • $\begingroup$ @jd27 ah thank you for adding that. Yes of course. $\endgroup$
    – Nick
    Jun 9, 2023 at 21:15

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