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I have come across an expression like this,

$$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a), $$

where I expected to find just $\delta(x-a)$. When I thought about it, though, I realised maybe... they are identical? Because both yield $1$ when integrating over a domain that contains $a$ and $0$ otherwise, so both distributions behave identically.

So, can I say $$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a) = \delta(x-a) $$ or can I not?

TeicDaun
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2 Answers2

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Not quite. Bear in mind that

$$\frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}} \Bigg|_{x=a} = \frac{2 f(a)}{2 \sqrt{ f(a)^2 }} = \frac{f(a)}{|f(a)|} = \begin{cases} 1 & f(a) > 0 \\ -1 & f(a) < 0\end{cases}$$

since $\sqrt{z^2} = |z|$. So the most you can claim is that

$$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a) = \begin{cases} \delta(x-a) & f(a) > 0 \\ -\delta(x-a) & f(a) < 0\end{cases}$$

PrincessEev
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  • In my case ---sorry I didn't mention---, $f(x) > 0$ for all $x$, so there is no problem with the absolute value resulting of the root-square cancellation. – TeicDaun Dec 02 '21 at 01:13
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    Ah I see. Then yeah, $\delta(x-a)$ is perfectly valid then – PrincessEev Dec 02 '21 at 01:51
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The famous property of sampling of the delta function yields $$ f(x)\delta(x-a)=f(a)\delta(x-a). $$ However, subtlety lies in this property. It should be mentioned that the validity of the above equation is the domain of $f(x)$, which in this case, is the set $\{x:f(x)f(a)>0\}$. Other than this, no more considerations are required.

Mostafa Ayaz
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