# Relations between Bochner spaces $L^p(0,T;L^q)$

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?

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.

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 $$u(t,x)=f(x)$$.