Almost, but not quite, an inner prouct

Question

Let $V$ be the vector space of continuous functions on a closed interval $[a, b]\subset \mathbb R$. Define a binary functional $\langle \, , \,\rangle: V \times V \to \mathbb R$ by $$\langle f, g \rangle = \sup \{|f(x)g(x)| \, \vert x \in [a,b]\}$$ This is not an inner product, because it is not bilinear. However, it is close! In particular it is:

1. Symmetric: $\langle f, g \rangle = \langle g, f \rangle$.
2. Positive definite: $\langle f, f \rangle \ge 0$ with equality if and only if $f = 0$.
3. Subadditive: $\langle f + g, h \rangle \le \langle f, h \rangle + \langle g, h \rangle $.
4. Homogeneous: $\langle cf, g \rangle = |c| \langle f, g \rangle$ for any $c \in \mathbb R$.

Note that if we could replace the inequality in property 3 with equality, and drop the absolute value in property 4, we would have an inner product.

Furthermore, this satisfies the Cauchy-Schwarz inequality: $\langle f, g \rangle^2 \le \langle f, f\rangle \langle g, g \rangle$. And, if that were not enough, if we try to define a norm in the usual way, via the definition $\| f \| = \sqrt{\langle f, f \rangle}$, we end up with precisely the uniform norm (which is usually offered as one of the simplest examples of a norm that is not induced by an inner product, because it does not obey the parallelogram law).

So clearly things like this are potentially quite useful. What are they called? Is this a "pseudo-inner product" (or should that be "inner pseudo-product"?) Or is "quasi-", or "semi-" the correct prefix to use? Do all mappings that satisfy 1-4 also satisfy Cauchy-Schwarz, or is this example special in that regard?

Answer 1

Some additional properties:

(4) implies

5. $\langle 0,g\rangle = 0$

(4) + (5) + (3) gives $$ 0 = \langle f - f,g\rangle \leq \langle f,g\rangle + \langle -f,g\rangle = 2 \langle f,g\rangle $$ and so we get

6. $\langle f,g\rangle \geq 0$ for all $f,g$.

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Additional Examples:

A large class of objects that satisfy your conditions 1-4 can be constructed in the following way. Fix $X$ some vector space. Consider its two-fold, symmetric tensor product $S$ (generated by elements of the form $x\otimes y + y \otimes x$ where $x,y\in X$). Fix a norm on $S$. Let $\langle x,y\rangle = \|x \otimes y + y \otimes x\|_S$. Then it is pretty easy to check that all 4 of the properties you prescribed hold true. (Properties 1 and 2 are obvious, 3 follows from triangle inequality of the norm, and 4 the scaling homogeneity of the norm.)

Now let $X = \mathbb{R}^2$ and $S$ can be identified with the symmetric $2\times 2$ matrices. A particular norm on $S$ is given by

$$ \left\| \begin{pmatrix} a & b \\ b & c \end{pmatrix} \right\| = |a| + 5|b| + |c| $$

The induced product on $X\times X$ is

$$\langle (x_1,x_2),(y_1,y_2)\rangle = 2|x_1 y_1| + 2|x_2 y_2| + 5 |x_1 y_2 + x_2 y_1| $$

Let $x = (0,1)$ and $y = (1,0)$, you find that $$ \langle x,x\rangle = 2 = \langle y,y\rangle $$ but $$ \langle x,y\rangle = 5 $$ and of course $5^2 \not\leq 4$, and so Cauchy-Schwarz is not immediate from 1 through 4.

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More Examples:

Another family of examples come from treating $X$ as an commutative algebra over the reals. Assume that $X$ also is normed as a vector space with norm $\|\text{-} \|_X$. Then we can define $$ \langle x,y\rangle = \|xy\|_X $$

Commutativity guarantees symmetry; triangle inequality for the norm guarantees the subadditivity. Scaling homogeneity of the norm also guarantees the same for the product.

The product now is automatically positive semi-definite, but if you assume further that $X$ has the property that $x^2 = 0 \implies x = 0$ then positive definiteness is guaranteed.

(This is the case of your question, where the space of continuous functions on a compact interval $[a,b]$ is a commutative Banach algebra with the uniform norm.)

In this case you also get examples showing that Cauchy-Schwarz is not always true.

Let $X = \mathbb{C}$ considered as a real algebra, with the usual complex multiplication. Choose my norm to be

$$ \| x + i y\| = |x| + 10|y| $$

Then

$$ \langle 1,i\rangle = \| i\| = 10 $$

But

$$ \langle 1,1\rangle = \|1\| = 1 = \|-1\| = \langle i,i\rangle $$

so your version of Cauchy-Schwarz fails.

(Note that the example given here is a perfectly good commutative Banach algebra: the standard Banach algebra version of Cauchy-Schwarz holds, in that $\|wz\| \leq \|w\|\|z\|$. The problem here is that $\|w^2\|$ can be a lot smaller than $\|w\|^2$. On the other hand, if you start with a commutative Banach algebra that has the property that $\|f^2\| = \|f\|^2$ for all $f$, then this construction will give you a product that also satisfies Cauchy-Schwarz.)