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I am reading Reinhold Remmert's book "Theory of Complex Functions" ...
I am focused on Chapter 0: Complex Numbers and Continuous Functions ... and in particular on Section 1.3: Scalar Product and Absolute Value ... ...
I need help in order to fully understand Remmert's derivation of the Cauchy-Schwartz Inequality ...
The start of Remmert's section on Scalar Product and Absolute Value reads as follows:
View attachment 8545
In the above text by Remmert we read the following:
" ... ... Routine calculations immediately reveal the identity
$$ \langle w, z \rangle^2 + \langle i w, z \rangle^2 = \mid w \mid^2 \mid z \mid^2$$ for all $$w,z \in \mathbb{C} $$
which contains as a special case the
Cauchy-Schwarz Inequality
$$\mid \langle w, z \rangle \mid \le \mid w \mid \mid z \mid$$ ... ...
... ... ... "
Now ... although it doesn't quite ft the language of a "special case" ...
... if $$\langle i w, z \rangle^2 \ge 0$$ ... then ... we could argue that ...$$\langle w, z \rangle^2 \le \mid w \mid^2 \mid z \mid^2$$ ... ...Is that the right move?... BUT ... can we be sure that ..
... $$\langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) \ge 0 $$ where we have $$w = u + iv$$ and $$z = x + iy$$ ...
... We have to worry that it might be the case that $$2uxvy \gt (v^2 x^2 + u^2 y^2 )$$ ...
... problem with suggested move above! ...Can some please clarify ... and indicate how to show that
$$\langle w, z \rangle^2 + \langle i w, z \rangle^2 = \mid w \mid^2 \mid z \mid^2$$
$$\Longrightarrow \mid \langle w, z \rangle \mid \le \mid w \mid \mid z \mid$$ ... ...
Hope that someone can help ...
Peter
I am focused on Chapter 0: Complex Numbers and Continuous Functions ... and in particular on Section 1.3: Scalar Product and Absolute Value ... ...
I need help in order to fully understand Remmert's derivation of the Cauchy-Schwartz Inequality ...
The start of Remmert's section on Scalar Product and Absolute Value reads as follows:
View attachment 8545
In the above text by Remmert we read the following:
" ... ... Routine calculations immediately reveal the identity
$$ \langle w, z \rangle^2 + \langle i w, z \rangle^2 = \mid w \mid^2 \mid z \mid^2$$ for all $$w,z \in \mathbb{C} $$
which contains as a special case the
Cauchy-Schwarz Inequality
$$\mid \langle w, z \rangle \mid \le \mid w \mid \mid z \mid$$ ... ...
... ... ... "
Now ... although it doesn't quite ft the language of a "special case" ...
... if $$\langle i w, z \rangle^2 \ge 0$$ ... then ... we could argue that ...$$\langle w, z \rangle^2 \le \mid w \mid^2 \mid z \mid^2$$ ... ...Is that the right move?... BUT ... can we be sure that ..
... $$\langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) \ge 0 $$ where we have $$w = u + iv$$ and $$z = x + iy$$ ...
... We have to worry that it might be the case that $$2uxvy \gt (v^2 x^2 + u^2 y^2 )$$ ...
... problem with suggested move above! ...Can some please clarify ... and indicate how to show that
$$\langle w, z \rangle^2 + \langle i w, z \rangle^2 = \mid w \mid^2 \mid z \mid^2$$
$$\Longrightarrow \mid \langle w, z \rangle \mid \le \mid w \mid \mid z \mid$$ ... ...
Hope that someone can help ...
Peter