# How to show functional derivative as a limit of ordinary derivative?

I found this footnote in the appendix (on path integral page 333) of J. Polchinski’s string theory book. can you explain this?

• Ultimately this might be better suited to mathematics, you might find a more convincing argument from a mathematical point of view instead. Oct 7, 2020 at 0:10
• Related and also. Oct 7, 2020 at 17:15

Recall the definition, $$\delta F[q;\phi]= \int \!\! dt~\frac{\delta F}{\delta q(t)} \phi(t) = \lim_{\varepsilon\to 0}\frac{F[q+\varepsilon \phi]-F[q]}{\varepsilon} = \left [ \frac{d}{d\varepsilon}F[q+\varepsilon \phi]\right ]_{\varepsilon=0},$$ where $$\phi (t)$$ specifies the direction of the functional derivative in $$\delta q(t)$$.
Now take a special $$F[q]\equiv \int \!\! d\tau ~~\delta(\tau-t') q(\tau)=q(t')$$, so $$\frac{\delta F}{\delta q(t)}= \delta(t-t').$$
It's easiest if you think of t as the continuum limit of a discrete integer index i, so a vector $$q_i \to q(t)$$. Note that the vector calculus gradient, $$\partial q_i/\partial q_j = \delta_{ij}$$ tends to the above expression.
Then, for example, the gradient of a scalar function goes to a vector, $$\frac{\partial (q_jq_j)}{\partial q_i} =2q_i ~~~\to ~~~\frac{\delta \int\!\!dt' ~q(t')q(t') }{\delta q(t)}= 2q(t).$$