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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.

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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 $|T(x,y)| \leq \|x \|y\|$.

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  • $\begingroup$ 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. :) $\endgroup$ – user38770 Apr 14 at 16:42
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    $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Apr 14 at 23:16

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