# Total derivative of integral seen as a functional, how?

1. Mar 27, 2012

### birulami

To be specific, with total derivative I mean the linear map that best approximates a given function $f$ at a given point. For $f:ℝ\toℝ$ we have $D(f,x_0):ℝ\toℝ$, i.e. $D(f,x_0)(h) \in ℝ$. Often it is also denoted as just $\delta f$.

Now in physics, in particular in the area of the Lagrangian, I find the following. Let $S_{a,b}(f) = \int_a^b f(x)dx$ a functional that maps functions $f$ to the real line. Then $D(S_{a,b},f) = \delta S_{a,b}$ should be well defined given any necessary smoothness conditions. In particular $D(S_{a,b},f)$ maps functions $h$ of the same type of $f$ to real numbers. Because the integral is linear, so my hunch, its best linear approximation should be itself. Yet in a physics course, equation 1.5, first line, I find what I understand to be

$$\delta \int_a^b f(x) dx = \int_a^b \delta f dx$$

Can anyone explain how the algebraic types on the left and on the right would match up? My interpretation is, that on the left I have a the total derivative of a functional, which itself should be a functional, written explicitly as $D(S_{a,b},f)$. On the right I have the integral over, hmm, the total derivative of $f$, where I don't see how this could be a functional?

Any hints appreciated.