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

Question

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 $\epsilon \delta(x-x) = 0$?

Answer 1

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 $\epsilon.$