Rudin 1.37 (d) is trivial to Rudin but hell for me

[josueortega]

Hi everyone,

I am working on my own through Rudin's Principles of Mathematical Analysis and, after the demonstration of Cauchy - Schwarz Inequality, in Theorem 1.37, part (d), Rudin states:

$$|x \cdot y| \leqslant |x||y|$$

When he explains how to prove this, he simply states that this is an immediate consequence of Schwarz Inequality, which he defines as follows:

$$|\sum_{j=1}^{n}a_j \overline{b_j}|^2 \leqslant \sum_{j=1}^{n}|a_j|^2 \sum_{j=1}^{n}|b_j|^2$$

If someone can explain me how this two things are identical I would appreciate it a lot. My toughts so far:

$$|x \cdot y| \leqslant |x||y|$$ Take the square of this, which is:

$$ (x \cdot y)(x \cdot y) \leqslant \sum x_i^2 \sum y_i^2$$ and hence,

$$ (\sum x_iy_i)^2 \leqslant \sum x_i^2 \sum y_i^2$$ which is NOT the Schwarz Inequality!

 

[Muphrid]

Sure it is? He just conjugates the $b_j$ to make the formula compatible with complex spaces. It's the same reason he slaps absolute value bars everywhere.

 

[josueortega]

I understand the part of the conjugate, actually the whole part on the left is pretty clear to me. The problem is the right side of the inequality. I don't understand how

$$\sum x_i^2 \sum y_i^2 = \sum_{j=1}^{n} |x|^2 \sum_{j=1}^{n} |y|^2$$

 

[Muphrid]

You mean $|x_i|$ on the right, not $|x|$, right? Again, it's just slapping absolute value bars around to make the inequality compatible with complex spaces. For real numbers, there is no difference.

 

[blue_raver22]

Hello,

I understand your confusion and I am happy to provide some clarification. The Schwarz Inequality that Rudin is referring to is also known as the Cauchy-Schwarz Inequality, which states:

$$|\sum_{j=1}^{n}a_j \overline{b_j}|^2 \leqslant \sum_{j=1}^{n}|a_j|^2 \sum_{j=1}^{n}|b_j|^2$$

This inequality is used in the proof of Theorem 1.37 (d), which states:

$$|x \cdot y| \leqslant |x||y|$$

To see how these two are related, let's take a closer look at the proof.

First, note that the dot product of two vectors $x$ and $y$ can be written as:

$$x \cdot y = \sum_{i=1}^{n} x_i y_i$$

Now, if we take the absolute value of both sides, we get:

$$|x \cdot y| = \left|\sum_{i=1}^{n} x_i y_i\right|$$

Using the Cauchy-Schwarz Inequality, we have:

$$\left|\sum_{i=1}^{n} x_i y_i\right| \leqslant \sqrt{\sum_{i=1}^{n}|x_i|^2}\sqrt{\sum_{i=1}^{n}|y_i|^2}$$

Note that the square root of the sum of squared terms is just the norm of a vector. So, we can rewrite this as:

$$|x \cdot y| \leqslant |x||y|$$

which is exactly what we wanted to prove.

In summary, the Schwarz Inequality is used in the proof of Theorem 1.37 (d) to show that the dot product of two vectors is bounded by the product of their norms. I hope this helps clarify things for you. Keep up the good work with Rudin's book!