I need help with this exercise:
Let X be a complex Banach space. A bilinear functional on $X\times X$ is a map $B: X \times X\rightarrow \mathbb{C}$ such that for all $x,y \in X$, the maps $B(x,\cdot),B(\cdot,y): X \rightarrow \mathbb{C}$ are both linear. Consider the product space $X\times X$ with the norm $\|(x,y)=\|(x,y)\|_{X\times X}=\|x\|+\|y\|$. By $(x_n,y_n)\rightarrow (x,y)$ we mean convergence in $X\times X$ in this norm. We say that B is jointly continuous at $(x,y)\in X\times X$ if $(x_n,y_n)\rightarrow(x,y)$ implies $B(x_n,y_n)\rightarrow B(x,y).$
a) Show that if $B: X\times X \rightarrow \mathbb{C}$ is jointly continous at $(0,0)$ then B is jointly continuous everywhere. (Hint: note that bilinearity in particular means that $B(x,y)=B(x/\sqrt{s},\sqrt{s}y)$, for $s >0$.)
b) Show that if B is jointly continuous everywhere, then there exists a constant $C>0$ such that $|B(x,y)|\le C\|x\|\|y\|$ for all $x,y \in X$.(Hint: Banach-Steinhaus).
I am struggling with a) but I think I managed b). Do you have any tips for a), and could you check if I solved b correct please?
For a), I tried this:
Let $(x_n,y_n) \rightarrow (x,y)$. $|B(x_n,y_n)-B(x,y)|=|B(x_n,y_n)-B(x,y_n)+B(x,y_n)-B(x,y)|=|B(x-x_n,y_n)+B(x,y_n-y)|$. But this doesn't seem to lead anywhere. I am also not quite sure how we are supposed to use the hint given.
For b) I tried:
$|B(x,y)|=|f_x(y)|$, where $f_x(\cdot)=B(x,\cdot)$. Now joint continuity gives continuity in one variable, so we get that:
$|f_x(y)|\le\|f_x\|\|y\|$. I will be able to solve the problem if I then can show that $\|f_x\|\le C\|x\|$. If x is the zero vector it is ok. If x is not the zero-vector, we look at the family of operators indexed by x, given by $f_x/\|x\|$. Then if we keep y, then for every operator given by x we have:
$|f_x(y)/\|x\||=|B(x,y)|/\|x\|\le \|f_y\|\|x\|/\|x\|=\|f_y\|$. So by the uniform boundedness principle, we have that $\|f_x\|/\|x\|\le C$. (Here we use that X is a Banach-space.)
Is b correct? And could you please give some hints for a)?
I find it easier to prove (b) first; using the hint that was given for (a) appears more relevant to (b).
(b) If no such $C$ exists, there is a sequence $(x_n,y_n)$ such that $$|B(x_n,y_n)| = 1\quad\text{ and }\quad \|x_n\|\|y_n\|\to 0 \tag{1}$$ By replacing $(x_n,y_n)$ with $(sx_n, s^{-1}y_n)$ we can arrange $\|x_n\|=\|y_n\|$ without changing $(1)$. Hence, $(x_n,y_n)\to 0$ and we have a contradiction with continuity at $(0,0)$.
(a) Using (b), the manipulations you tried lead to the desired conclusion: $$ |B(x_n,y_n)-B(x,y)|=|B(x_n,y_n)-B(x,y_n)+B(x,y_n)-B(x,y)| \\ =|B(x-x_n,y_n)+B(x,y_n-y)| \\ \le C(\|x-x_n\| \|y_n\|+\|x\| \|y_n-y\|)\to 0 $$