Use Cauchy-Schwartz inequality to prove that $ : mathscr H times mathscr H to Bbb C$ is continuous.($mathscr H$ is a Hilbert Space.)

Let $(a,b) in mathscr H times mathscr H$ be fixed.

So we have to prove that for a given $epsilon gt 0$, we can find $delta_1 gt 0$ and $delta_2 gt 0$ such that $|<x,y>-<a,b>| lt epsilon$ whenever $||x-a|| lt delta_1$ and $||y-b||lt delta_2$ and to do this we have to use The Cauchy-Schwartz inequality $|<x,y>| le ||x||||y||$.

(Note that “$<,>$” denotes the inner product and “$||cdot||$” denotes the norm induced by it.)

Let $epsilon gt 0$ be given.

Then,

I start with $|<x,y>-<a,b>|=|<x,y>-<a,b>+<x,b>-<x,b>|=|<x,y-b>+<x-a,b>| le |<x,y-b>|+|<x-a,b>| le ||x||||y-b||+||x-a||||b||.$

Now I choose $delta_1=frac {epsilon}{||b||}$ and $delta_2=frac {epsilon}{||a||}$.

Hence,$||x||-||a|| le||x-a|| lt delta_1$ and $||y-b||lt delta_2$

$Rightarrow$

$|<x,y>-<a,b>| le (||a||+delta_1)delta_2 + delta_1 ||b||=(||a||+frac {epsilon}{||b||})frac {epsilon}{||a||}+epsilon=2epsilon + frac {epsilon^2}{||a||||b||}.$

So we conclude that $<,>$ is continuous.

Am I on right track so far? Is there a more simpler way to do this (Using Cauchy-Schwartz inequality obviously.)?