First I know that this question has all ready be asked for exemple here but for bilinear operator with only one variable here I want to show it for linear operator with two variables.
Question:
Prove that a bilinear operator $\phi$ is continuous iff verifying $|| \phi (v;w)|| \leq M ||v|| ||w|| $
Answer:
1 - $\forall v \in E, w \in F, || \phi (v;w)||_G \leq M ||v||_E ||w||_F \Rightarrow \phi $ is continuous
By definition $ \phi(.) $ is continuous iff $\forall$ couple of sequence $ (v_n,w_n) \in E \times F$ verifying $\lim_{n \to \infty}(v_n;w_n) = (v;w) $ we have that $\lim_{n \to \infty} \phi (v_n;w_n) = (v;w)$
$0 \leq || \phi(v;w) - \phi(v_n;w_n)||_G = || \phi(v-v_n;w) - \phi(v;w - w_n)||_G \leq || \phi(v-v_n;w)||_G + || \phi(v_n;w - w_n)||_G \leq M ||v-v_n ||_E ||w ||_F + M ||w-w_n ||_F ||v ||_E \xrightarrow[n \to \infty]{} 0$
When the limit is obtain by simple definition of limit arithmetic. And thus by sandwitch theorem Q.E.D.
2- $\phi $ is continuous $ \Rightarrow $ $\forall v \in E, w \in F, || \phi (v;w)||_G \leq M ||v||_E ||w||_F$
On one hand if $ \phi$ is continuous we have that for $\epsilon = 1 , \exists \delta > 0 : \forall (v;w) \in B_E(0;\delta) \times B_F(0;\delta) \Rightarrow || \phi(v;w) ||_G < 1$
On an other hand we can write $ || \phi(v;w) ||_G = || \phi(v \frac{\delta}{||v||} \frac{||v||}{\delta};w \frac{\delta}{||w||} \frac{||w||}{\delta} ) ||_G =
|| \frac{||v||}{\delta} \cdot \frac{||w||}{\delta} \cdot \phi(v \frac{\delta}{||v||} ;w \frac{\delta}{||w||} ) ||_G < 1 \cdot \frac{||v||}{\delta} \cdot \frac{||w||}{\delta}$
To sum up we have that $ || \phi(v;w) ||_G < 1 \cdot \frac{||v||}{\delta} \cdot \frac{||w||}{\delta} $
Q.E.D.
Is this correct? Mostly the demonstration in "1-" as it is different from other demonstration in 1D using Lipshchitz to prove continuity. But any feed back on '2-' will be greatly appreciated ( as I am studiying alone).
Thank you.
