Define $L_{t}^{s}L_{x}^{q}$ to be the set of functions defined on the interval $(0,T)$ valued in $L_{x}^{q}$, such that
$$\int_{0}^{T} \lVert v(t,.)\rVert_{q}^s dt<\infty,$$
where $t$ and $x$ indicate corresponding variables. I would like to show that if $v\in L_{t}^{\infty}L_{x}^{2}\cap L_{t}^{2}L_{x}^{6} $, then $v\in L_{t}^{s}L_{x}^{q}$ for $\frac{3}{q}+\frac{2}{s}=\frac{3}{2}$ and $q\in [2,6]$. The hint is that using H$\ddot{o}$lder inequality first in $x$ and then in $t$. Many thanks for any suggestions and solutions.
So let me write a more precise solution. First note that for any $1\leq p\leq q\leq r\leq \infty$, defining $\theta$ such that $\frac{1}{q} = \frac{1-\theta}{p} + \frac{\theta}{r}$), by Hölder's inequality $$ \int |v|^q = \int |v|^{q(1-\theta)} |v|^{q\theta} \leq \|v\|_{L^p}^{q\,(1-\theta)} \, \|v\|_{L^r}^{q\,\theta} $$ since $\frac{q(1-\theta)}{p} + \frac{q\theta}{r} = 1$. This gives what is also called sometimes an interpolation inequality. $$ \|v\|_{L^q} \leq \|v\|_{L^p}^{1-\theta} \, \|v\|_{L^r}^{\theta}. $$ In particular, in your case, since $2\leq q \leq 6$, $$ \|v\|_{L^q} \leq \|v\|_{L^2}^{1-\theta} \, \|v\|_{L^6}^{\theta}. $$ with $\theta = \frac{3}{2} - \frac{3}{q}$. Hence $$ \int_0^T\|v\|_{L^q}^s\,\mathrm d t \leq \|v\|_{L^\infty_tL^2_x}^{s(1-\theta)}\int_0^T \, \|v\|_{L^6}^{s\theta}\,\mathrm d t. $$ In particular, choosing $\theta$ such that $s\theta = 2$ is equivalent to choose $q$ such that $s \,(\frac{3}{2} - \frac{3}{q})= 2 \iff \frac{3}{q} + \frac{2}{s} = \frac{3}{2}$.