I'd been struggling with this problem:
Prove that a bilinear functional $B \in \mathcal{B}(D \times D, \mathbb{R})$ is continuous iff there exist a positive number K such that
$$|B[x,y]| \leq K \| x \|\, \| y \| $$
Where $x,y \in D$ (remeber D is the set of Admissible functions)
And the norm is defined by: $\|(x,y)\| = \max \{\|x\|, \|y\|\}$
My try: If I suppose that: $|B[x,y]| \leq K \,\| x\|\,\|y\|$, then this implies that the $B$ is continous, because it is bounded, Is it right?
But I don't know how to approach the other implication
I will try to give you some ideas without the full solution. So your exercise is to show that continuous and bounded are equivalent, so you have the right idea, but you have to demonstrate the equivalence, not just invoke it.
Two hints: It might be helpful for you to refer to the proof that linear operators (on a normed space) are continuous if and only if they are bounded. The machinery of that proof essentially works identically for the bilinear case, since bilinear maps are linear maps when one of the arguments is held constant.
The other hint is: can you prove that the real-valued function $f(x,y) = axy$ is continuous, where $a$ is any real number? The function $f$ is just a bilinear map on a 1-dimensional vector space.
If you need a more explicit lead, we should show two things: (1) bounded implies continuous, and (2) not bounded implies not continuous. By referring to my above hint, (1) can be proven. To approach (2), we can see that not bounded means that there exists a sequence $(x_n, y_n)$ such that $B(x_n, y_n) > n\|x\| \|y\|$. Then, use the bilinearity of $B$ to rescale $(x_n, y_n)$ so that you get that a sequence of points going to $0$, but such that the value of $B$ does not go to $0$.