Bilinear function bounded by norms of arguments Let $v_1 \in X$ and $v_2 \in Y$, $v = (v_1, v_2) \in X\times Y$, $X$ and $Y$ being normed linear spaces. Let $\phi$ be a continuous bilinear function.
Why
$$\|\phi(v_1, v_2)\| \leq K\cdot\|v_1\|\cdot\|v_2\|\quad ?$$
If the function is continuous, I can understand something of the type
$$\|\phi(v_1, v_2)\|\cdot\frac1{\|v\|} \leq K$$
But I'm having trouble getting from one expression to the other.
Thanks
 A: Hint: What can we say about the map $x\mapsto\,\big(y\mapsto \phi(x,y)\big)$? 
A: Answer:
Since $f$ is continuous bilinear, we have that $f$ is continuous at 0. 
Then, for an open ball $B_\mathbb{R}(0, \epsilon)$, we have that $f^{-1}(B_\mathbb{R}(0, \varepsilon))$ is open in $X$ x $Y$. (continuity implies that the pre-image of an open set is open).
Then, we deduce the existence of $\delta > 0 $ such that:
$||x||<\delta \Rightarrow ||f(x)||<\varepsilon$
Then, since X x Y is a linear space, we have that $x = (z, y) \in X$ x $Y$; then, $\delta z \in X$ and $\delta y \in Y$. Then $\delta (z,y)$ = $\delta x$ belongs to the cartesian product of $X$ x $Y$
$||\delta*x|| < \delta \Rightarrow ||f(\delta*x)|| \varepsilon$
We then have that 
$\delta ||x|| < \delta \Rightarrow ||\delta^2f(x)|| \varepsilon$
This follows from bilinearity (take delta out of each argument of the function, each at a time, by fixing the other argument).
Thus, if $||x||<1$, we have that $||f(x)|| < \frac{\varepsilon}{\delta^2}$ 
Then, $||f||^* = sup\{||f(x)||: ||x||<1\} < \infty$
Now consider $f(x, y)$ = $||x||*||y||*f(x^*,y^*)$ where $x^*$ = $\frac{x}{||x||}$ (this is a standard operation)
Ok, now we have the conclusion, since
$||x||||y||f(x^*,y^*) \leq ||x||*||y||*||f||^*$
You can name $||f||^*$ as "K" and get the desired result.
Kudos to my math TA.
