Exploring Remmert's Derivation of Cauchy-Schwartz Inequality

In summary: Alternatively, we can write $\langle w,z\rangle^2 = (vx-uy)^2 \leqslant (v^2+u^2)(x^2+y^2) = |w|^2|z|^2$, so again taking the positive square root gives the desired result.
  • #1
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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

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

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  • #2
Peter said:
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, \(\displaystyle \langle i w, z \rangle^2 = (v^2 x^2 - 2uxvy + u^2 y^2 ) = (vx-uy)^2 \geqslant 0 \).
 
  • #3
Opalg said:
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
 
  • #4
Opalg said:
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
 
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  • #5
Peter said:
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 $|\langle w,z\rangle| \leqslant |w||z|$.
 

FAQ: Exploring Remmert's Derivation of Cauchy-Schwartz Inequality

What is Cauchy-Schwartz inequality?

Cauchy-Schwartz inequality, also known as Cauchy-Bunyakovsky-Schwarz inequality, is a mathematical inequality that states the relationship between the inner product of two vectors and the norms of those vectors. It states that the absolute value of the inner product is less than or equal to the product of the norms of the two vectors.

Who derived the Cauchy-Schwartz inequality?

The Cauchy-Schwartz inequality was derived by Augustin-Louis Cauchy, a French mathematician, in the 19th century. However, it was also independently derived by Viktor Yakovlevich Bunyakovsky and Hermann Amandus Schwarz in the same time period.

What is Remmert's derivation of Cauchy-Schwartz inequality?

Remmert's derivation of Cauchy-Schwartz inequality is a proof that uses the concept of projections and Pythagorean theorem to derive the inequality. It involves using geometric interpretations and algebraic manipulations to prove the inequality.

Why is Cauchy-Schwartz inequality important?

The Cauchy-Schwartz inequality is important because it has many applications in various fields of mathematics, such as linear algebra, functional analysis, and probability theory. It also serves as a fundamental tool in proving other mathematical theorems and inequalities.

What are some real-life examples of Cauchy-Schwartz inequality?

Cauchy-Schwartz inequality can be seen in various real-life situations, such as calculating the maximum distance a person can travel in a given time with a fixed speed, determining the shortest possible length of a diagonal in a rectangle, and finding the maximum area of a rectangle with a fixed perimeter. It also has applications in economics, physics, and engineering.

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