# Why is there an “absolute value” and a norm in the Schwarz Inequality?

This really bothers me, and I'm not sure if it's just that I'm not understanding it correctly. For the moment, assume we are working in a vector space $V$ over $\mathbb{R}^n$. Let $x,y \in V$. We have an inner product $\langle \cdot , \cdot\rangle\colon V \to \mathbb{R}$, which in this case is just the dot product.

In so many presentations of the Schwarz Inequality, the inequality is given as:

$\big\rvert\langle x,y\rangle\big\lvert \leq \big\rvert\big\rvert{x}\big\lvert\big\lvert \;\big\rvert\big\rvert{y}\big\lvert\big\lvert$

Why do we use both the absolute value and the norm? The absolute value is a norm, so why don't we just write

$\big\rvert\big\rvert \langle x,y\rangle\big\lvert\big\lvert^2 \leq \big\rvert\big\rvert{x}\big\lvert\big\lvert \;\big\rvert\big\rvert{y}\big\lvert\big\lvert$

I'm not sure how clear this question is, but I guess my gripe is with the fact that we present the inequality with two different types of norms.

I guess, I find it easier to conceptually understand the following notation:

$\langle x,y \rangle^2 \leq \langle x, x \rangle \langle y, y \rangle$

This is actually really bothering me at the moment, and I can't figure out why the "absolute value" is such a popular way to present the inequality.

• They are norms on two different spaces. The absolute value on the left is a norm on $\mathbb{R}$ (or $\mathbb{C}$), whereas the norm on the right is a norm on $\mathbb{R}^n$ (or some more general inner product space). So I think it's reasonable to use different notation. The usual convention is to use $| \cdot|$ for absolute value on $\mathbb{R}$ or $\mathbb{C}$, and $\|\cdot\|$ for norms on all other spaces. – Nate Eldredge May 23 '14 at 5:29
• By the way, in your second displayed equation, you don't want the square on the left side. – Nate Eldredge May 23 '14 at 5:30
• Note that just leaving out the absolute value, $\langle x,y\rangle \le \left\|x\right\|\left\|y\right\|$, is still equivalent to the full claim – Hagen von Eitzen May 23 '14 at 5:39
• @NateEldredge Isn't the square necessary? $\big(\sum{x}_i{y}_i\big)^2 \leq \sum{x}_i^2 + \sum{y}_i^2$ Sorry if I'm being dense here. – angryandconfused May 23 '14 at 5:42
• No, the square should not be there (or if it is then you also need to square the other side). Inspect the proof more carefully. – Nate Eldredge May 23 '14 at 5:57

I guess my gripe is with the fact that we present the inequality with two different types of norms.

That's because they are two different norms, on two different spaces! Moreover, we might not even want to think of the absolute value as being a norm for this purpose.

When you declare a normed space $(V, \lVert{-}\rVert)$, you're assigning the symbol $\lVert{-}\rVert$ as your notation for the norm on that space. If $V \ne \mathbb{R}$, say $V = \mathbb{R}^3$ or whatever, then $\lVert{-}\rVert$ refers to the norm on $\mathbb{R}^3$ and not on $\mathbb{R}$.

However, the notation $|{-}|$ unambiguously means one thing: the absolute value.

• Thanks, that helped a little. I guess the fact that they are two different norms makes it seem, well... $meh...$ It just seems confusing to think about it that way. Something seems less 'perfect.' I don't know... I need to think about it some more. – angryandconfused May 23 '14 at 5:50

The norm operates on the vector space that contain $x$ and $y$. The absolute value operates on the scalar. Moreover, the absolute value has a very precise meaning in this case: removal of the sign of real numbers. However, the norm (in this case derived from the dot product) can be based on any of the infinitely many dot products in existence. It's really a very specific metric that can be constructed on vector spaces with a dot products (=square root of the dot product with itself). It's true that the absolute value is a norm... but not any norm is a dot product.

I too prefer the last expression. It's clearer and doesn't introduce square roots, which are indeed a very specific construct. What if the vectors were constructed over the field of integers? In that case, the square root doesn't necessarily exist, but a square is still well defined.