When dealing with time-dependent PDEs, one often obtain that some quantity $E(t,x)$ belongs to a Lebesgue space $L^p_t(L^q_y)$, which means that $$\int_0^{+\infty}\|E(t,\cdot)\|_{L^q(\mathbb{R}^n)}^p dt<+\infty.$$ Sometimes, we even have $E\in L^{p_1}_t(L^{q_1}_y)\cap L^{p_2}_t(L^{q_2}_y)$ ; when this occurs, a simple interpolation, which involves only Hölder inequality, yields $E\in L^p_t(L^q_y)$ where $(\frac1p,\frac1q)$ is any point in the segment between $(\frac1{p_1},\frac1{q_1})$ and $(\frac1{p_2},\frac1{q_2})$.

I am interested in a slightly different situation, where $E\in L^{p_1}_t(L^{q_1}_y)\cap L^{q_2}_y(L^{p_2}_t)$ (the order of integrations differ in both spaces). What are the interpolation spaces ?

For instance, suppose that $E\in L^\infty_t(L^1_y)\cap L^\infty_y(L^1_t)$. What can be said of $E$ ?

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    This old answer on Math.SE might contain some bits of information. – Giuseppe Negro Nov 12 at 11:12
  • @GiuseppeNegro. But it does not answer my (rather specific) question. – Denis Serre Nov 12 at 11:21
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    No, it does not. I was just hoping it could be a bit useful. Sorry if it was not. – Giuseppe Negro Nov 12 at 11:23
  • There is something not (completely) trivial that can be said by interpolation: namely that for every $f \in L^2_t$, the function $y \mapsto \int E(t,y) f(t) dt$ belongs to $L^2_y$. And conversely by replacing $t$ and $y$. In other words, $E$ is the kernel of a bounded operator from $L^2_y$ to $L^2_t$. Indeed, the assumption $E \in L^\infty_t(L^1_y)$ is equivalent to $E$ being bounded $L^1_t \to L^1_y$ and conversely. – Mikael de la Salle Nov 16 at 15:10

$E \in L^\infty_t(L^1_y) \cap L^\infty_y (L^1_t)$ is insufficient to give any information about whether $E$ belongs to $L^p_t L^q_y$ when $p < \infty$.

Example: for simplicity, let's work with sequence spaces. Let $$ E(t,y) = \delta_{t,y} \, .$$ Then clearly $E \in \ell^\infty_t \ell^1_y \cap \ell^\infty_y \ell^1_t$. But $E$ doesn't belong in any $\ell^p_t \ell^q_y$ space where $p < \infty$.

  • If either $p_1 \leq q_1$ or $q_2 \leq p_2$ you can use Minkowski to swap the order, and the standard result applies. When $p_1 > q_1$ and $q_2 > p_2$, modifying the above shows that you cannot embed into $\ell^p \ell^q$ with $p < \mathrm{min}(p_1, q_2)$. – Willie Wong Nov 12 at 16:23
  • Your example doesn't work, it is only $\ell^1\ell^1$. But you can take the measure $\delta_D$ supported by the diagonal $D=\{t=y\}$. – Denis Serre Nov 12 at 17:43
  • @DenisSerre: my example is not $\ell^1, \ell^1$. ($\delta$ is the Kronecker, since we are working in sequence spaces). So $\sum_{y} E(t,y) = 1$ for every $t$ and vice versa. – Willie Wong Nov 12 at 20:59
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    @Amir: thanks for fixing the typo! – Willie Wong Nov 13 at 6:25

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