Is $\langle f \rangle$ an “inner product”?

Let $$\langle f(x,y)\rangle = \iint_S f(x,y)\,\mathrm{d}x\,\mathrm{d}y$$

I have seen the above in multiple papers as the definition of $\langle f(x,y)\rangle$. I would normally associate angle brackets with being an inner product $\langle f,g \rangle$ of two functions $f$ and $g$, but the definition I have quoted appears to be acting on a single function. Is this still defining $\langle f(x,y) \rangle$ as an inner product, or is it borrowing the notation of angle brackets resulting in my confusion?

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An inner product takes two inputs, this takes only $f$. – Thomas Andrews Mar 29 '13 at 1:24
What I have seen is $\langle f, g\rangle = \int fg d\mu.$ – ncmathsadist Mar 29 '13 at 1:25
This is a linear functional. – Isaac Solomon Mar 29 '13 at 1:30
In many contexts $\langle A \rangle$ refers to some sort of average or expectation value of an object $A$, and the definition you've quoted is suggestive of this. Indeed, if $S$ has unit area, then $\langle f(x,y) \rangle$ really is the average value of $f$ on $S$. – Branimir Ćaćić Mar 29 '13 at 1:36
It's definitely not an inner product. As Branimir says, it appears to mean something like the expected value of the function $f$. – Twiceler Mar 29 '13 at 2:06

Often $\langle f \rangle_\mathcal{F}$ denotes the average or expected value of $f$ over the family/region $\mathcal{F}$.
In my experience, it's typically normalized (but not always) when that makes sense. So if $\mathcal{F}$ is a discrete family, then $\langle f \rangle_\mathcal{F}$ might typically be defined as $$\langle f \rangle_\mathcal{F} = \frac{1}{\# \{x \in \mathcal{F} \}} \sum f(x).$$ If $\mathcal{F}$ is continuous, then it might typically be defined as $$\langle f \rangle_\mathcal{F} = \frac{1}{|\mathcal F|} \int_\mathcal{F} f d \mu.$$ When the domain $\mathcal{F}$ is understood, it is often left from notation.