Total derivative of integral seen as a functional, how?

birulami
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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.
 
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When it comes to derivatives, the notations are even more varying than the perspectives of a derivative itself. I think you put too much interpretation into the ##\delta## notation. The LHS is the derivative of a real number? Doesn't make sense, but I don't see the functional either. I read it as ##\delta F = \int f'## where ##F## denotes the antiderivative of ##f## and ##f'## its derivative. Hence it is a version of the fundamental theorem of calculus written in another way. As mentioned, do not expect consistency in notation of derivatives.

https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
 

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