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$E_1,E_2,F$ are Banach spaces. There is one thing I don't understand if the following proof:

enter image description here

I don't understand how he can say "This proves the first assertion."

I'm assuming that $E_1 \times E_2$ is provided the sum norm $\|(h_1,h_2)\| = \|h_1\| + \|h_2\|$, so if we call $\lambda : E_1 \times E_2 \to \,\, ; \,\, (h_1,h_2) \mapsto \omega(x_1,h_2) + \omega(h_1,x_2)$, what he shows is that

$\frac{\omega((x_1,x_2) + (h_1,h_2)) - \omega(x_1,x_2) - \lambda(h_1,h_2)}{\| (h_1,h_2) \|} \\ = \frac{ \omega(x_1+h_1,x_2+h_2) - \omega(x_1,x_2) - (\omega(x_1,h_2) + \omega(h_1,x_2))}{ \|(h_1,h_2) \| } \\ = \frac{ \omega(h_1,h_2) }{ \| (h_1,h_2) \|} = \frac{ \omega(h_1,h_2)}{ \|h_1\| + \|h_2\|}$

due to bilinearity.

I know that $\| \omega(h_1,h_2) \| \leqslant \| \omega \| \|h_1\| \|h_2 \|$, where

$\| \omega \| = sup_{\|h_1\|=1,\|h_2\|=1} \omega(h_1,h_2)$

but I don't think that is enough to prove that $\omega(h_1,h_2)/(\|h_1\|+\|h_2\|) \to 0$ as $(h_1,h_2) \to 0$, since I don't think

$\frac{\|h_1\| \|h_2\| }{\|h_1\| + \|h_2\|}$

tends to zero as $(h_1,h_2) \to 0$. What am I missing?

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It tends to $0$. Recall that $$ \frac{\|h_1\|}{\|h_1 \| + \|h_2\|} \le 1 $$ hence $$ \frac{\|h_1\|\|h_2\|}{\|h_1 \| + \|h_2\|} \le \|h_2\| \to 0, \qquad (h_1,h_2) \to 0 $$

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  • $\begingroup$ Wow big oversight on my part. Thank you for your help $\endgroup$ – joeb Jan 21 '17 at 18:43

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