Exploring Remmert's Derivation of Cauchy-Schwartz Inequality

[Math Amateur]

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

\(\displaystyle \langle w, z \rangle^2 + \langle i w, z \rangle^2 = \mid w \mid^2 \mid z \mid^2\) for all \(\displaystyle w,z \in \mathbb{C} \)

which contains as a special case the

Cauchy-Schwarz Inequality

\(\displaystyle \mid \langle w, z \rangle \mid \le \mid w \mid \mid z \mid\) ... ...

... ... ... "
Now ... although it doesn't quite ft the language of a "special case" ...

... if \(\displaystyle \langle i w, z \rangle^2 \ge 0\) ... then ... we could argue that ...\(\displaystyle \langle w, z \rangle^2 \le \mid w \mid^2 \mid z \mid^2\) ... ...Is that the right move?... BUT ... can we be sure that ..

... \(\displaystyle \langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) \ge 0 \) where we have \(\displaystyle w = u + iv\) and \(\displaystyle z = x + iy\) ...

... We have to worry that it might be the case that \(\displaystyle 2uxvy \gt (v^2 x^2 + u^2 y^2 )\) ...

... problem with suggested move above! ...Can some please clarify ... and indicate how to show that

\(\displaystyle \langle w, z \rangle^2 + \langle i w, z \rangle^2 = \mid w \mid^2 \mid z \mid^2\)

\(\displaystyle \Longrightarrow \mid \langle w, z \rangle \mid \le \mid w \mid \mid z \mid\) ... ...
Hope that someone can help ...

Peter

 

[Opalg]

can we be sure that ..

... \(\displaystyle \langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) \ge 0 \) where we have \(\displaystyle w = u + iv\) and \(\displaystyle z = x + iy\) ...

... We have to worry that it might be the case that \(\displaystyle 2uxvy \gt (v^2 x^2 + u^2 y^2 )\) ...

The inner product $\langle w,z\rangle$ is by definition real, so its square must necessarily be positive (or zero). In fact, \(\displaystyle \langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) = (vx-uy)^2 \geqslant 0 \).

 

[Math Amateur]

The inner product $\langle w,z\rangle$ is by definition real, so its square must necessarily be positive (or zero). In fact, \(\displaystyle \langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) = (vx-uy)^2 \geqslant 0 \).

Thanks for the help, Opalg ...

Peter

 

[Math Amateur]

The inner product $\langle w,z\rangle$ is by definition real, so its square must necessarily be positive (or zero). In fact, \(\displaystyle \langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) = (vx-uy)^2 \geqslant 0 \).

I have just become aware that the argument:

\(\displaystyle \langle w, z \rangle^2 + \langle i w, z \rangle^2 = \ \mid w \mid^2 \mid z \mid^2\) for all \(\displaystyle w,z \in \mathbb{C} \)

\(\displaystyle \Longrightarrow\) \(\displaystyle \langle w, z \rangle^2 \le \ \mid w \mid^2 \mid z \mid^2\)

\(\displaystyle \Longrightarrow\) \(\displaystyle \langle w, z \rangle \ \le \ \mid w \mid \mid z \mid\)does not show that \(\displaystyle \mid \langle w, z \rangle \mid \ \le \ \mid w \mid \mid z \mid \)

... since

... ... \(\displaystyle \mid \langle w, z \rangle \mid \ \le \ \mid w \mid \mid z \mid\)

\(\displaystyle \Longrightarrow - \mid w \mid \mid z \mid \ \le \ \mid \langle w, z \rangle \mid \ \le \ \mid w \mid \mid z \mid\)... so ... effectively we have only shown "half" of the Cauchy-Schwarz Inequality ...Can someone please demonstrate a full proof of the inequality ... preferably demonstrating that \(\displaystyle - \mid w \mid \mid z \mid \ \le \ \mid \langle w, z \rangle \mid\) ...Peter

 

[Opalg]

I have just become aware that the argument:

\(\displaystyle \langle w, z \rangle^2 + \langle i w, z \rangle^2 = \ \mid w \mid^2 \mid z \mid^2\) for all \(\displaystyle w,z \in \mathbb{C} \)

\(\displaystyle \Longrightarrow\) \(\displaystyle \langle w, z \rangle^2 \le \ \mid w \mid^2 \mid z \mid^2\)

\(\displaystyle \Longrightarrow\) \(\displaystyle \langle w, z \rangle \ \le \ \mid w \mid \mid z \mid\)does not show that \(\displaystyle \mid \langle w, z \rangle \mid \ \le \ \mid w \mid \mid z \mid \)

... since

... ... \(\displaystyle \mid \langle w, z \rangle \mid \ \le \ \mid w \mid \mid z \mid\)

\(\displaystyle \Longrightarrow - \mid w \mid \mid z \mid \ \le \ \mid \langle w, z \rangle \mid \ \le \ \mid w \mid \mid z \mid\)... so ... effectively we have only shown "half" of the Cauchy-Schwarz Inequality ...Can someone please demonstrate a full proof of the inequality ... preferably demonstrating that \(\displaystyle - \mid w \mid \mid z \mid \ \le \ \mid \langle w, z \rangle \mid\) ...

If $a$ and $b$ are real numbers, then $a^2\leqslant b^2$ always implies $|a|\leqslant |b|$ (because $(-a)^2 = a^2 = |a|^2$).

In this case, if $\langle w,z\rangle^2 \leqslant |w|^2|z|^2$, then taking the positive square root of both sides it follows that $|\langle w,z\rangle| \leqslant |w||z|$.