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Good day!

Let $V = H^1(\Omega)$, $\Omega \subset \mathbb R^3$.

Consider the space $W = \{ y \in L^2(0,T;V) \colon dy/dt \in L^2(0,T;V') \}$.

It is well-known that $W \subset C([0,T];H)$ where $H = L^2(\Omega)$.

My question: for what $\alpha$ and $\beta$ we may assert that $W \subset L^\alpha(0,T;L^\beta(\Omega))$?

For example, $W \subset L^\infty(0,T;L^2(\Omega))$ and $W \subset L^2(0,T;L^6(\Omega))$ because $H^1(\Omega) \subset L^6(\Omega)$.

May we assert that $W \subset L^8(0,T;L^{24/5}(\Omega))$?

Thank you!

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    $\begingroup$ You have already noted that the two end-points are $L^\infty_t L^2_x$ and $L^2_t L^6_x$. It remains to interpolate between the two. If I did the computations right, the interpolant is either $L^{16/7}_t L^{24/5}_x$ or $L^8_{t} L^{12/5}_x$, and so you may not assert your final line. $\endgroup$
    – Willie Wong
    Commented May 9, 2014 at 8:31
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    $\begingroup$ Basically solving the interpolation you get $1/\beta + 2/(3\alpha) = 1/2$ as the sharp interpolant. For fixed $\beta$ you can reduce $\alpha$ as $[0,T]$ has finite measure. If $\Omega$ is bounded then you can also reduce $\beta$ for fixed $\alpha$. Increasing, however, is not allowed. $\endgroup$
    – Willie Wong
    Commented May 9, 2014 at 8:37

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This is an interpolation problem. $$ L^\infty(0,T;L^2(\Omega))\cap L^2(0,T;L^6(\Omega)) \hookrightarrow L^{2/\theta}(0,T;L^{1/[(1-\theta)/2+\theta/6]}(\Omega)) , $$
for any $\theta\in(0,1)$. You can check that this space coincides with $L^2(0,T;L^6(\Omega))$ when $\theta=1$, and this space coincides with $L^\infty(0,T;L^2(\Omega))$ when $\theta=0$.

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  • $\begingroup$ This type of interpolation method can be found in the book "Interpolation spaces: an introduction", written by J. Bergh and J. Lofstrom. $\endgroup$
    – Buyang LI
    Commented May 12, 2014 at 13:31

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