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Let $(X,\|\cdot\|_X)$ be a Banach space. The function $u=u(t,x)$ belongs to Bochner space $L^p(0,T;X)$ if the norm $$ \|u\|_{L^p(0,T;X)} = \left(\int_0^T \|u(t,\cdot)\|_X^p \mathrm{d}t\right)^{1/p} \text{ for } p<\infty, $$ $$ \|u\|_{L^\infty(0,T;X)} = \mathrm{ess}\sup_{t\in[0,T]} \|u(t,\cdot)\|_X $$ is finite.

Now let $X=L^q(\Omega)$ for bounded $\Omega\subset\mathbb{R}^n$. I was wondering, if there are some embeddings between spaces $L^{p_1}(0,T;L^{q_1}(\Omega))$ and $L^{p_2}(0,T;L^{q_2}(\Omega))$, depending on $p_i$ and $q_i$.

From the interpolation in Lebesgue spaces we have this result, moreover from the boundedness of $[0,T]$ and $\Omega$ we obviously have $$L^p(0,T;L^{q_1}(\Omega)) \subset L^p(0,T;L^{q_2}(\Omega)) \text{ for } q_1>q_2$$ and $$ L^{p_1}(0,T;L^q(\Omega)) \subset L^{p_2}(0,T;L^q(\Omega)) \textrm{ for } p_1>p_2. $$

However, is it possible that $$ L^{p_1}(0,T;L^{q_1}(\Omega))\subset L^{p_2}(0,T;L^{q_2}(\Omega)) $$ for some $p_1>p_2$ and $q_1<q_2$?

In other words, can we compensate the integrability over one variable by the higher integrability of the other? In particular I am interested in the inclusion of $L^2(0,T;L^2(\Omega))$ and $L^1(0,T;L^\infty(\Omega))$.

EDIT: it seems that there's no relation whatsoever. In particular if $\Omega=[0,T]=[0,1]$ there are simple examples for no such relation between $L^2(0,T;L^2(\Omega))$ and $L^1(0,T;L^\infty(\Omega))$: take $f(t,x)=\frac{x}{\sqrt{t}}$ and $g(t,x)=\frac{t}{x^{1/4}}$ - one of them is in $L^1(0,T;L^\infty)$ but not in $L^2$ and the second one vice versa. I believe these examples can be easily modified to other cases.

I'm not deleting this post in case somebody would have a similar problem in the future.

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No, this is not true, as long as $L^{q_1}(\Omega)\setminus L^{q_2}(\Omega)\neq\emptyset$.

You can see it considering a function $f\in L^{q_1}(\Omega)\setminus L^{q_2}(\Omega)$ and setting $u(t,x)=f(x)$.

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