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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.

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    $\begingroup$ 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 $\int_0^1 \int_0^1 |f(x,y)|^2 dx dy = 0$. $\endgroup$ – Bungo Apr 20 '15 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 Apr 20 '15 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 $\int_0^1 \int_0^1 |f(x,y)|^2 dx dy = 0$. $\endgroup$ – Bungo Apr 20 '15 at 17:22
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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.

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