Let $V$ be a hilbert space and $a(.,.)$ is a continuous, coercive bilinear map on $V*V$.
Then how to calculate the derivative of $a(.,.)$ and if $$J(u)=\frac{1}{2}a(u,u)$$ then how to show $(J'(u),v)=a(u,v)$.
I do have some idea of calculating the derivative of $a(u,v)$ as $$\frac{a(u+\epsilon h,v+\epsilon k)-a(u,v)}{\epsilon}=\frac{a(u+\epsilon h,v+\epsilon k)-a(u,v+\epsilon k)}{\epsilon}+\frac{a(u,v+\epsilon k)-a(u,v)}{\epsilon}$$ and passing to the limit on both the sides we get R.H.S. =$2a(u,v)$
I am not sure about the proof and i do not have any idea for $J'$. Any type of help will be appreciated. Thanks in advance.
Let $Da(x) \in L(V\times V , V)$ denote the derivative of $a$ at the point $x=(x_1,x_2)\in V \times V$. Then for some direction $h=(h_1, h_2)\in V \times V$ we have (if the derivative exists) $$ Da(x)h=\frac{d}{d\varepsilon}_{\varepsilon = 0}a(x+\varepsilon h).$$ Let's try to calculate this using the bilinearity of $a$: $$ a(x+\varepsilon h) = a (x_1+\varepsilon h_1 , x_2 + \varepsilon h_2) = a(x_1 , x_2) + \varepsilon a(x_1 , h_2) + \varepsilon a(h_1 , x_2) + \varepsilon^2a(h_1,h_2).$$ Taking the derivative with respect to $\varepsilon$ is now very easy: $$ \frac{d}{d\varepsilon}a(x+\varepsilon h)=a(x_1,h_2)+a(h_1,x_2)+2\varepsilon a(h_1,h_2). $$ Setting $\varepsilon = 0$ we finally get $$ Da(x)h = a(x_1,h_2)+a(h_1,x_2). $$ Since by definition of the derivative the linear map $Da(x)$ should be continuous, $a$ has to be continuous as well. Notice that we didn't use the fact that $a$ is coercive anywhere. For the derivative of $J$, can you go ahead as in the first part?