Does $p$ integrability in n-1 dimensions give higher integrability in $n$ dimensions? Restrict everything to a ball in $n$ dimensions, let $x$ represent the first $n-1$ variables, and $t$ the $n-$th variable. It is obvious by Holder's Inequality that
$$
\int\limits_t\left(\int\limits_x|f|^p\right)^{\frac1p}\leq\left(\int\limits_t\int\limits_x|f|^p\right)^{\frac1p}|\{f\neq0\}|^{1-\frac1p}
$$
I am interested in whether the following inequality holds
$$
\int\limits_t\left(\int\limits_x|f|^p\right)^{\frac1p}\leq C\left(\int\limits_t\int\limits_x|f|^{p-\epsilon}\right)^{\frac1{p-\epsilon}}
$$
for $\epsilon>0$ and $f$ such that $\Vert f\Vert_{L^p(x)}$ and $\Vert f\Vert_{L^p(x,t)}$ are small ($<1$ in particular). Any results related to this question are welcome.
 A: This can't possibly hold. You are on a bounded set, so just take your $f(t,x) = g(x)$ for any function $g$ that is in $L^{p-\epsilon}$ but not $L^p$. (Take a sequence $\hat{f}_n \in L^p\cap L^{p-\epsilon}$ such that $\hat{f}_n \to g$ in $L^{p-\epsilon}$. Let $f_n = \delta \hat{f}_n / \| \hat{f}_n \|_{L^p}$. Then $f_n \to 0$ in $L^{p-\epsilon}$ showing that the uniform bound you are looking for is not possible. [Incidentally, since your inequality is linear, there is no point in restricting to "$f$ with sufficiently small norm".)
The reverse inequality cannot hold either by considering functions that are constant in the $x$ variable. 

The estimate could hold if you know something more about $f$, something that effectively controls a higher $L^q$ norm in the $x$ direction. For example, if you know regularity you can appeal to Sobolev. You can probably also get by with something that is weaker than $L^q$: just something about scaling behavior on small sets should be enough. 
Alternatively, you may also be able to recover the inequality if $x$ and $t$ are not truly independent variables. If $f$ has some hidden correlation between the $x$ and $t$ directions, then it may be possible to do what you want, though typically I would expect the inequality to run the other way.
