# Which averages of products of a function give a norm?

Let $$f: [0,1] \rightarrow \mathbb{R}$$ be a bounded measurable function. For some real non-negative numbers $$a_1, a_2, b_1, b_2$$ with $$a_1+b_1=a_2+b_2=1$$ consider the quantity $$N(f)=\int_{[0,1]} \int_{[0,1]} f(a_1x+b_1y)f(a_2x+b_2y) \, dx \, dy.$$

If $$a_1=a_2=1$$ and $$b_1=b_2=0$$ then $$N(f)^{1/2}$$ is just the $$L_2$$ norm of $$f$$. I am interested to understand when the quantity $$N(f)^{1/2}$$ defines a norm in other situations. Is anything like that known? Maybe at least some clear necessary conditions for the thing to be a norm?

Naturally, if this is known and easy, the general question would for a $$k$$-wise product of $$f$$ with some affine combination of the integrating variables plugged in each $$f$$ and then the question would be whether or not $$N(f)^{1/k}$$ is a norm.

• Does in the assumption $f : [0,1] \to [0,1]$ the restriction to $[0,1]$-valued functions make any sense? I think you need at least some vector space of functions. – Dieter Kadelka Mar 27 at 16:52
• Thank you, indeed, I will correct the conditions. – TOM Mar 28 at 2:11

## 1 Answer

It is a norm when $$a_1=a_2$$, $$b_1=b_2$$. Otherwise making the change of variables $$u=a_1x+b_1y, v=a_2x+b_2y$$ we get the integral of $$f(u)f(v)$$ over certain parallelogram $$P$$ with a diagonal joining $$(0, 0)$$ and $$(1, 1)$$. Assume that $$f$$ has a small support $$\Delta$$ so that $$\Delta^2\subset P$$. Then the integral is just $$(\int f)^2$$ and this may be equal to 0.

• Thank you for the answer! Just one more question in case it is easy - would anything significant change if I assume that f is non-negative? – TOM Mar 28 at 14:38
• but the norm should be defined on a linear space, not on a cone? – Fedor Petrov Mar 28 at 15:50
• Sorry, I am being silly, I should think more before immediately asking. Thank you for your clear answer once again! – TOM Mar 29 at 12:51