Let $f\in L^2[0,1]^2$. Does it follow that $$\int_0^1|f(x,x)|dx<\infty\,?$$

By the Cauchy-Schwarz inequality, $$\int_0^1|f(x,x)|dx\leq \left(\int_0^1|f(x,x)|^2dx\right)^{1/2} = \left(\int_0^1\int_0^1|f(x,x)|^2dxdy\right)^{1/2}.$$

So I need to change variables somehow.

  • 4
    $\begingroup$ The set $D = \{(x,x) | x \in [0,1]\}$ has measure zero as a subset of $[0,1]^2$. We can assign any value we like to $f$ on this set without affecting $\int_0^1 \int_0^1 |f(x,y)|^2 dx dy$. For example, put $f(x,y) = \infty$ for all $(x,y) \in D$, and $f(x,y) = 0$ otherwise. Then $\int_0^1 |f(x,x)|dx = \infty$ but $\int_0^1 \int_0^1 |f(x,y)|^2 dx dy = 0$. $\endgroup$
    – user169852
    Commented Apr 20, 2015 at 16:58
  • $\begingroup$ Nice observation. What if we assume that $f(x,y)<\infty,\forall x,y$, would something like this come up as a counterexample? $\endgroup$
    – Aad
    Commented Apr 20, 2015 at 17:20
  • $\begingroup$ Instead of setting $f(x,x) = \infty$, we can use any function which is not integrable on $[0,1]$. For example, $f(x,x) = 1/x$ for $x \in (0,1]$, and $f(x,y) = 0$ everywhere else. Then the same result holds: $\int_0^1 |f(x,x)| dx = \infty$ but $\int_0^1 \int_0^1 |f(x,y)|^2 dx dy = 0$. $\endgroup$
    – user169852
    Commented Apr 20, 2015 at 17:22

1 Answer 1


No, the value $\int_0^1 |f(x,x)|\,dx$ is not even well defined for $f \in L^2([0,1]^2)$.

Recall that the elements of $L^2([0,1]^2)$ are strictly speaking not functions on $[0,1]^2$; they are equivalence classes of functions, mod equality almost everywhere. As Bungo's comment says, if you let $D = \{(x,x) : x \in [0,1]\}$ be the diagonal of the square, then $D$ has Lebesgue measure zero. So the functions $f_1 = 0$ and $f_2 = 1_D$ represent the same element of $L^2([0,1]^2)$, but they have different values for $\int_0^1 |f(x,x)|\,dx$. We could likewise choose representatives for which the integral on the diagonal was infinite, or nonexistent.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .