In this widely cited and wildly popular proof of the Cauchy-Schwarz inequality, the authors write (http://www.math.lsa.umich.edu/~speyer/417/CauchySchwartz.pdf)

Let $u$ and $v$ be two vectors in $R^n$. The Cauchy-Schwartz inequality states that $$|u \cdot v| ≤ |u||v|$$

I go to another source (Proof that the Euclidean norm is indeed a norm), a comment states states:

Cauchy-Schwarz: $u \cdot v \le \lVert u\rVert \lVert v\rVert$

Yet another source (http://rgmia.org/papers/v12e/Cauchy-Schwarzinequality.pdf):

$$\alpha \cdot \beta = |\alpha| |\beta| \cos (\alpha, \beta)$$ we deduce that $$\alpha \cdot \beta \le |\alpha| |\beta|$$

Is Cauchy Schwarz inequality

Lastly, On Wikipedia:

$$|\langle x,y\rangle| \leq \|x\| \cdot \|y\|$$

I must say I am slightly disappointed in the notational inconsistency in the literature.

What is the rule/logic for applying the absolute value sign and the norm when it comes to the Cauchy-Schwarz inequality? What is the most correct way to express the CS inequality?

  • $\begingroup$ The Cauchy-Schwarz inequality should have absolute value signs on the left:$| \langle x, y \rangle | \leq \| x \| \| y \|$. $\endgroup$
    – littleO
    Dec 24 '15 at 23:30
  • 2
    $\begingroup$ To add one other comment the answers below haven't yet mentioned: the notation you learn at different stages of mathematical education are designed to make it easy to learn. The last version you have in your question is closest to the "mature" notation which generalizes to an wide and important class of vector spaces beyond $\mathbb R^n$. The earlier notation is generally restricted to use only in $\mathbb R^n$ and is designed to echo in part your understanding of the absolute value of numbers. $\endgroup$
    – Simon S
    Dec 24 '15 at 23:40

There are three separate issues:

  1. Using $\|v\|$ or just $|v|$ to express the norm of a vector.

  2. Using an absolute value on $\langle u, v \rangle$ or not.

  3. Using $\langle u, v \rangle$ or $u \cdot{} \ v$.

The first is really a matter of style. What speaks in favor of $\| v\|$ is to highlight that it is the norm of a vector, what speaks in favor of $|v|$ is that that $|u\dot \ v| \le |u| \ |v|$ makes the formulation really slick, perhaps overly so.

Also the third is just a matter of style. Both mean the (or a) scalar product of $u$ and $v$.

For the second, for real vectors spaces in a strict sense it is weaker without the absolute value, yet the one with the absolute value can be derived easily from the one without (considering the inequality with $-u$ instead). Thus it is basically the same. For complex vectors though, it does not make sense to write an inequality without the absolute value.

In any case, the content is that the absolute value of the scalar product of two vectors $u$ and $v$ is bounded above by the product of the norms (induced by the scalar product) of the two vectors.

  • $\begingroup$ Best answer here. Just to point out a version of the C-S inequality with/without absolute value is actually equivalent to the other. $\endgroup$
    – Simon S
    Dec 24 '15 at 23:35
  • 1
    $\begingroup$ It's equivalent provided you're working on $\mathbb{R}$. On the other hand, working on $\mathbb{C}$, you need the absolute value / modulus sign or else it doesnt make sense. $\endgroup$
    – Batman
    Dec 25 '15 at 3:18
  • $\begingroup$ @Batman that's a good point. I was mostly thinking about the reals, as this is what is assumed in most sources linked. I will add a remark though. $\endgroup$
    – quid
    Dec 25 '15 at 11:35

Give a real or complex vector space $V$ and a positive semidefinite symmetric sesquilinear form $\langle \cdot, \cdot \rangle$ on $V$ one can define the norm of of an element $v \in V$ as $\|v\| = \sqrt{\langle v, v \rangle}$. The Cauchy-Schwarz inequality then states that $$ |\langle v, w \rangle| \leq \|v\| \cdot \|w\| \quad \text{for all $v,w \in V$}. $$ Depending on the situation one can now use slightly different notations or state this in slightly different, but equivalent ways:

Instead of writing $\langle \cdot, \cdot \rangle$ or $\|\cdot\|$ one might prefer to use a simple $\cdot$ to denote the sesquilinear form and $|\cdot|$ to denote the norm. This is mostly a matter of taste, but depending on the vector space $V$ the notations $|\cdot|$ and $\cdot$ may already be occupied. Take for example the real vector space $C([0,1])$ of the continuous realvalued functions on the unit interval, together with the scalar product $$ \langle f,g \rangle = \int_0^1 f(t) g(t) \,\text{d}t. $$ Then the notation $f \cdot g$ is already occupied with the pointwise product of $f$ and $g$ and $|f|$ with the pointwise absolute value of $f$.

Regarding the left side of the inequality the first thing to realize is that in the case of a complex vector space $V$ the scalar $\langle v, w \rangle$ is a complex number and not necessarily real. But the inequality $$ \langle v, w \rangle \leq \|v\| \cdot \|w\| $$ does not make sense if $\langle v, w \rangle$ is not real. So in this case one can not omit the absolute value.

If we are working over the real numbers, then it is possible to omit the absolute value from the left side of the inequality without weaking the inequality. To see this notice that $x \leq |x|$ for all $x \in \mathbb{R}$, so it follows from the Cauchy-Schwarz inequality that $$ \langle v, w \rangle \leq |\langle v, w \rangle| \leq \|v\| \cdot \|w\| \quad \text{for all $v,w \in V$}. $$ Interestingly the reverse also follows, because for every $v,w \in V$ we then have \begin{gather*} \langle v, w \rangle \leq \|v\| \cdot \|w\| \quad\text{and}\quad -\langle v, w \rangle = \langle -v, w \rangle \leq \|-v\| \cdot \|w\| = \|v\| \cdot \|w\|, \end{gather*} and thus $ |\langle v, w \rangle| = \max\{\langle v,w \rangle, -\langle v, w \rangle\} \leq \|v\| \cdot \|w\|. $


It's very simple. Whenever you have a real inner product space and define the norm of $x$ to be the square root of the scalar product of $x$ with itself, then you have the Cauchy-Schwarz inequality. Whatever is the notation for the inner product and for the norm.

Remark 1: The formula with the $\cos$ may be understood as the definition of the angle.

Remark 2: The C-S inequality $$\mathrm{|scal\_product|} \leq \mathrm{product\_of\_norms}$$ immediately implies $$\mathrm{scal\_product} \leq \mathrm{product\_of\_norms}$$

  • $\begingroup$ It seems more relevant (and slightly less obvious) that the latter inequality also implies the former. $\endgroup$
    – quid
    Dec 24 '15 at 23:34
  • $\begingroup$ Yes, they are obviously equivalent, but it seems to me that the name CS inequality usually refers to the former. $\endgroup$ Dec 24 '15 at 23:36
  • $\begingroup$ I agree on what is common. I now see what you point was. $\endgroup$
    – quid
    Dec 24 '15 at 23:39

Sometimes, one likes to differiantiate between the norm of a vector $v$ and the modulus of a real $a$. In this case, you'd write $\|v\|$ and $|a|$.

The wikipedia article uses this notation ($|\langle x,y\rangle|\leq\|x\|\|y\|$). This is the notation I view as the best and clearest.

The middle source omits the modulus of the scalar product (since $\langle x,y\rangle \leq |\langle x,y\rangle|$) and can afford to drop the extra notation for the norm - the only norm involved are the vector norms $|x|,|y|$: $\langle x,y\rangle \leq|x||y|$. You could use $\|x\|\|y\|$ here as well, but this would be a waste of ink.

And finally, the first reference burdens the reader with the task to find out which norm is actually referred to: The norm $|u\cdot v|$ on the left hand side refers to the modulus of a real, the norms $|u|, |y|$ on the r.h.s. refer to the vector norms. This saves some ink at the expense of the reader's synapses.

Conclusion: To use the same symbol $|\cdot|$ for different things is sloppy, but acceptable if the readers know which instance of the symbol is meant. (In terms of programming, you could call $|\cdot|$ an overloaded function.)


I find the other expressions better than $\alpha\cdot\beta\leq |\alpha||\beta|$ since $\alpha\cdot\beta$ might be negative and if it is negative it does not assure $-\alpha\cdot\beta\leq |\alpha||\beta|$. But looking at $\alpha\cdot\beta=|\alpha||\beta|\cos(\alpha,\beta)$ one knows that $$ |\alpha\cdot\beta|\leq |\alpha||\beta| $$ which is a stricter conclusion, just as the one you had from the article from umich.edu.

In the Wikipedia article they use bracket notation to perform dot product and then take its absolute value. Also they take, what may appear to be, the dot product between two scalars.

My way of trying to be as pedantic as possible would be to write $$ |\alpha\cdot\beta|\leq ||\alpha||_2\,||\beta||_2 $$ to assure we take the Eucliadian norm, i.e. $\mathcal l_2$ $norm$. Here we use $|\cdot|$ for scalars and $||\cdot||_2$ for vectors.


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