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I am studying "Partial Differential Equation. An introduction." by Walter Strauss and at the beginning of Ch. 12, while talking about distributions, it says the delta function "is not a function. It's a more general object called a distribution. A function is a rule that assigns numbers to numbers. A distribution is a rule (or transformation, or functional) that assigns numbers to functions."

It goes on to say: "The delta function is the rule that assigns the number f(0) to the function f(x)."

Can someone clarify that last statement?

I understand that f(x) is referred to as a "test function"; what is the definition of a test function?

Thanks.

Federico Gremmo
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  • SeeTHIS ANSWER and THIS ONE for primers on The Dirac Delta. – Mark Viola Oct 26 '16 at 20:55
  • And a "test function" is simply any function belonging to a specified space of functions. The space of functions that are $C^\infty$ and have compact support are considered a standard space of test functions and gives rise to standard distributions. The space of smooth test functions that are rapidly decreasing faster than any polynomial increases gives rise to tempered distributions. – Mark Viola Oct 26 '16 at 21:07

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$$\lim_{\alpha\to 0}\int_{-\infty}^{\infty}\delta(x,\alpha)f(x)dx = \int_{-\infty }^{\infty}f(x)\delta (x) dx = f(0)$$ take this as a definition to avoid calculation For example : $$ let \ \delta(x,\alpha)=\frac{1}{\alpha\sqrt{2\pi}}e^\frac{-x^{2}}{2\alpha^{2}} $$ \begin{align} \lim_{\alpha\to 0}\int_{-\infty}^{\infty}\delta(x,\alpha)\, f(x) \, dx &= \int_{-\infty }^{\infty}f(x)\delta (x) dx = f(0) \\ &=\lim_{\alpha\to 0}\int_{-\infty}^{\infty}\frac{1}{\alpha\sqrt{2\pi}}e^\frac{-x^{2}}{2\alpha^{2}}f(x)dx\\&= \lim_{\alpha\to 0}\int_{-\infty}^{\infty}\frac{1}{\sqrt{\pi}}e^{-y^{2}}f(\sqrt{2}\alpha y)dy\\&=\lim_{\alpha\to 0}f(\sqrt{2}\alpha \xi)\int_{-\infty}^{\infty}\frac{1}{\sqrt{\pi}}e^{-y^{2}}dy\\&=f(0) \end{align}

Mahmoud Hassan
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