In the book about calculus of variation sand optimal control theory by Liberzon, he gives the following Taylor expansion in order to define the variations of a functional $J$:
However, for a normal $\mathbb R \to \mathbb R$ function, the Taylor expansion would have a $\frac 1 2 $ inserted there:
$$f(y+\alpha)=f(y)+f'(y)\alpha+\frac 1 2f''(y)\alpha ^2 + o(\alpha ^2)$$
Why is the $\frac 1 2$ not also next to the second-variation for the functional Taylor expansion?
I think that the result of this is that $$\delta ^2J|_y(\eta)=\frac 1 2\lim_{\alpha \to \infty}\left ( \frac{\delta J|_{y+\alpha \eta}(\eta)-\delta J|_y(\eta)}{\alpha}\right)$$
In other words, the $\frac 1 2 $ shows up here, where it shouldn't be according to my intuition.
Why is this?
EDIT: per request of @P.Siehr:
There is no particular reason that there is not a $1/2$ there; it's just a convention from the author (though perhaps an unconventional one).
I could do the same thing for a real-valued function $f$, and define the $n$th "variations" of $f$ to be the functions $f^{[n]}$ by $$f(x+\epsilon)=f(x)+f^{[1]}(x)\epsilon+f^{[2]}(x)\epsilon^2+\dots+f^{[n]}(x)\epsilon^n+o(\epsilon^n).$$ In this case Taylor's theorem tells us that $f^{[n]}(x)=f^{(n)}(x)/n!$, where $f^{(n)}$ indicates the $n$th derivative, so there is no need to introduce the new notation $f^{[n]}$.
That being said, I would guess the reason for omitting the $1/2$ in this case is that we don't care about the value of the second variation, only its sign. The usefulness of the second variation comes from looking at whether it is positive- or negative-definite, and an errant factor of $1/2$ doesn't affect that.