# Functional derivative for $J[f]=\int [f(y)]^p \phi(y)dy$

In QFT for gifted amateur pg. 13, the functional derivative for the functional

$$J[f]=\int [f(y)]^p \phi(y)dy$$

is given by

$$\frac{\delta J[f]}{\delta f(x)}= \lim_{\epsilon\rightarrow0} \frac{1}{\epsilon} (\int [f(y)+\epsilon\delta(y-x)]^p \phi(y) dy - \int[f(y)]^p \phi(y)) =p[f(x)]^{p-1} \phi(x).$$

To check that this is true, I expanded the first integral on the RHS,

$$\lim_{\epsilon\rightarrow0} \frac{1}{\epsilon} \int [f(y)+\epsilon\delta(y-x)]^p \phi(y) dy=\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int [f(y)^p+pf(y)^{p-1}\epsilon\delta(y-x) + \binom{p}{2}f(y)^{p-2}\epsilon^2\delta(y-x)^2+...]dy.$$

I have trouble showing that the integral $$\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int\binom{p}{2}f(y)^{p-2}\epsilon^2\delta(y-x)^2dy=0.$$

Carrying out the integration using one of the Dirac delta function, we get

$$\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int\binom{p}{2}f(y)^{p-2}\epsilon^2\delta(y-x)^2dy= \epsilon\binom{p}{2}f(x)^{p-2} \delta(x-x).$$

Since $$\delta(x-x)$$ is infinity, how can we say that this expression is zero? It seems to me that it is undefined since $$\epsilon$$ is also a very small number. Why can we say $$\epsilon \delta(x-x) = 0$$?

• This might be of interest. Feb 1 at 9:09
• Feb 1 at 10:44
• Why one should complicate her/his life evaluating a functional derivative by using an increment of the argument of the functional made by a Dirac delta function, instead of using a simpler (and more regular) form of the increment? Feb 1 at 13:07
• The term has an $\epsilon^2$ so it does not contribute to the derivative, which only keeps the $O(\epsilon)$ term Feb 1 at 13:25
• Related post by OP: physics.stackexchange.com/q/692240/2451 Feb 2 at 6:43

Since $$\delta(x)$$ is not an operational function, it can only be defined by a limiting process. Your variation is in fact including two limiting processes. For multi-limiting processes, the order of taking limitation is relevant to the final result.
In your formulation, you may replace the $$\epsilon \delta(x-y)$$ by a Gaussian function $$\epsilon \delta(x-y) = \lim_{\sigma\to 0}\frac{\epsilon}{\sigma\sqrt{\pi}} e^{-\left(\frac{x}{\sigma}\right)^2}$$ It means that you impose small a Gaussian deviation from the original function at near $$x$$ with a strength $$\epsilon$$. Then, the limit of $$\sigma$$ is performed after the limit of $$\epsilon.$$
• Also why must we take the limit of $\epsilon$ first instead of the limit of $\sigma$ first? Feb 2 at 4:04
• As for this problem, consider that you are doing a numerical computation for the functional variation. You put a small deviation (strength $\epsilon$) about a point $x_0$, perform the integration, find out $\Delta J (\epsilon)$ as function of the strength, take the limit $\epsilon \to 0$. After all this had done, you then may narrow the deviation to a $\delta$ function. If you limit you deviation into a $\delta$ function prior to the process mentioned above. The numerical process will not able to proceed using a $\delta$ function.