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Referring to this question: https://math.stackexchange.com/q/1355854 which was in turn a question about: $a: E\times F\to G$ bilinear separately continuous implies continuous?

I don't understand where $$\left\|\dfrac{a_x}{\|x\|}\right\|≤c$$ comes from, still.

The theorem merely states that $\|a_x y\|\leq c\|y\|$ for all $x\in E$, $y\in F$.

Kenny Wong
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Matthew
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1 Answers1

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Let me write it out more explicitly. Define a linear mapping $T : E \to \mathcal B(F, G)$ by $$ x \mapsto a_x,$$ where $a_x \in \mathcal B(F, G)$ is defined by $$ a_x(y) = a(x,y).$$ Notice that $a_x$ is indeed in $\mathcal B(F, G)$, since the mapping $y \mapsto a(x,y)$ is continuous for each $x \in E$.

Let $B(1) \subset E$ be the unit ball in $E$. We are going to show that $T(B(1))$ is a bounded subset of $\mathcal B(F, G)$.

By the uniform boundedness principle, it is only necessary to show that for every $y \in F$, the set $$\{ T(x)( y ) : x \in B(1) \} = \{ a(x,y) : || x || \leq 1 \}$$ is a bounded subset in $G$. This statement is true, because the mapping $x \mapsto a(x,y)$ is continuous for every $y \in F$.

Having shown that $T(B(1))$ is bounded, we deduce that there exists a $c$ such that $$ || x || \leq 1 \implies ||a_x || \leq c.$$ Therefore, $$ || a_x || \leq c ||x ||$$ for all $x \in E$.

Finally, $$ || a(x,y) || \leq ||a_x || ||y || \leq c || x || \ || y ||.$$

Kenny Wong
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  • P.S. I'm a bit worried because it seems as if we only used that $F$ is Banach (for the uniform boundedness principle). I can't see where we used the fact that $E$ and $G$ are Banach. I hope I didn't screw up... – Kenny Wong Mar 19 '17 at 22:58
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    Pretty sure only one of them needs to be Banach as I've seen versions of the question with only one of them being so. With regards to your solution, why do you restrict yourself to the unit ball and if you do so does the solution still hold for all x,y? – Matthew Mar 19 '17 at 23:06
  • I restrict myself to the unit ball because I can show that $T(B(1))$ is bounded in $\mathcal B(F,G)$ but I can't show that $T(E)$ is bounded in $\mathcal B(F,G)$. – Kenny Wong Mar 19 '17 at 23:07
  • The statement that $T(B(1))$ is bounded is the statement that $|| a_x || \leq c$ for some constant $c$, for all $x \in B(1)$. But then, by rescaling, you learn that $|| a_x || \leq c ||x ||$ for arbitrary $x \in E$. – Kenny Wong Mar 19 '17 at 23:08
  • ... and the inequality $|| a_x |\ \leq c || x|| $ is actually the inequality you want, not $|| a_x || \leq c$, which was written in one of the links you posted. – Kenny Wong Mar 19 '17 at 23:09
  • thanks, all clear now, very much appreciated – Matthew Mar 19 '17 at 23:12
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    The rescaling argument was confusing to me -- for a future person reading this. I reasoned as follows: We have for $x$ in a ball that -- $|x| \leq 1 \implies |a_x|\leq c$. So for any $x \in X$ for the implication to hold we have $|\frac{x}{|x|}| \leq 1 \implies |a_{\frac{x}{|x|}}|\leq c$. To see more clearly what this means write: $a_{\frac{x}{|x|}} = a(\frac{x}{|x|},\cdot) \in \mathcal{B}(F,G)$ so that $a_{\frac{x}{|x|}} = a(\frac{x}{|x|},\cdot) = \frac{1}{|x|}a(x,\cdot) = \frac{1}{|x|}a_x$... – yoshi Nov 28 '18 at 16:17
  • ...Thus the last part of the second implication will read, for all $x \in X$ we have $|\frac{a_x}{|x|}|\leq c$. Rearrange to finish. – yoshi Nov 28 '18 at 16:17