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Let $E_1, E_2$ and $F$ be $3$ Banach spaces. Let $B: E = E_1 \times E_2 \to F$ be a continuous bilinear form. Show that $B$ is differentiable at every point $a = (a_1,a_2) \in E$ and its differential is: $$dB_{(a_1,a_2)}(x_1,x_2) = B(x_1,a_2)+B(a_1,x_2)$$

We need to show that: $$B(x_1,x_2)-B(a_1,a_2) - [B(x_1,a_2)+B(a_1,x_2)] = o(||x-a||_E)$$ The LHS should be in some form of $B(x_1-a_1,x_2-a_2)$, then as $B$ is a continuous bilinear form, there exists $C > 0$ such that: $B(x_1-a_1,x_2-a_2) \leq C ||x_1-a_1||_{E_1} ||x_2-a_2||_{E_2} \leq C \displaystyle ||x_-a_||^2_E$, if we define the norm on $E$ by $||x = (x_1,x_2)||_E = \max(||x_1||_{E_1}, ||x_2||_{E_2})$.

However, when I calculate the LHS, it gives: $$B(x_1,x_2)-B(a_1,a_2) - [B(x_1,a_2)+B(a_1,x_2)] = B(x_1-a_1,x_2) - B(x_1+a_1, a_2) \text{ (?!)}$$ Where did I get wrong?

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Well, i think that you can try better with this expression: $B(x_1+a_1, x_2+a_2)-B(x_1,x_2)$.

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