# Functional which is linear and continuous in each variable is linear and bounded in both.

I am having trouble interpreting the following question:

Let $$\{X, || \cdot ||_{X} \}$$ and $$\{Y, || \cdot ||_{Y} \}$$ be Banach spaces. Let $$T(x,y) \colon X \times Y \to \mathbb{R}$$ be a functional linear and continuous in each of the two variables. Then $$T$$ is linear and bounded with respect to both variables.

Does showing that T is linear with respect to both variables mean showing that $$T$$ is bilinear? If so then it seems that it is by definition. If linearity is actually meant then I don't understand how $$T$$ can be linear in both coordinates, for example $$T(ax,ay) = a^2T(x,y) \neq aT(x,y)$$?

Assuming bilinear is meant I believe the question wants me to use a proposition which states that a family pointwise equi-bounded maps from a Banach space into a normed space are uniformly equi-bounded. It seems the problem would be finished if I could show one of the families $$\{ T(\cdot, y) \}_{y \in Y}$$ or $$\{ T(x, \cdot) \}_{x \in X}$$ is pointwise equi-bounded. However I don't see any reason why this would be true. My other thought is that maybe there is a way to use the Closed Graph Theorem since the question assumes both X and Y are Banach. However, I think we would need $$T$$ to be linear instead of bilinear to use its continuity to imply boundedness. Any hints or clarifications on the statement of the problem would be much appreciated.

What we can say is that $$T$$ is bilinear and bounded in the sense Thus $$|T(x,y)| \leq \|x \|y\|$$. For each $$y \in Y$$ define $$T_y$$ by $$T_y(x)=T(x,y)$$. $$T_y$$ is a bounded linear functional on $$X$$. Note that $$|T_y (x)|=|T(x,y)|\leq M_y \|x\|$$ for some finite constant $$M_y$$. Hence Uniform Boundedness Principle can be applied to conclude that $$|T_y(x)| \leq C\|x\|$$ for some $$C$$ independent of $$y$$. Thus $$|T(x,y)| \leq \|x \|y\|$$.
• Is the Uniform Boundedness Principle applied to the family $\{ T_{x} \}_{||x||=1}$, using that for $y$ fixed and for all $x$, $||x|| = 1$, $T_{x}(y) = T_{y}(x) \leq ||x|| M_{y} = M_{y}$? Also, do you mean $|T(x,y)| \leq ||x||y||$ up to multiplication by $C$? Thank you very much for your solution. :) – user38770 Apr 14 '19 at 16:42
• @user38770 What you are saying is correct but I am considering $(T_y)$ whereas you are considering $(T_x)$. Makes no differerence . Both are correct. – Kavi Rama Murthy Apr 14 '19 at 23:16