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.

  • $\begingroup$ 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. $\endgroup$ – Dieter Kadelka Mar 27 at 16:52
  • $\begingroup$ Thank you, indeed, I will correct the conditions. $\endgroup$ – TOM Mar 28 at 2:11

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.

  • $\begingroup$ 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? $\endgroup$ – TOM Mar 28 at 14:38
  • $\begingroup$ but the norm should be defined on a linear space, not on a cone? $\endgroup$ – Fedor Petrov Mar 28 at 15:50
  • $\begingroup$ Sorry, I am being silly, I should think more before immediately asking. Thank you for your clear answer once again! $\endgroup$ – TOM Mar 29 at 12:51

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