I have a somewhat easy looking question on parabolic function spaces:


Let $B$ be a ball in $\mathbb R^n$ and let $T>0$. Denote $Q:=B \times [0,T]$. Assume $f \in L^2(Q) \cap L^\infty(0,T; L^q(B))$ for every $1 \le q <2$. Does this imply that $f \in L^\infty(0, T; L^2(B))$?


I tried to build a counterexample, but at least my first attempt failed. I decided to ask here, because this is probably very easy for an expert. In addition, I would be interested in knowing what might be a good reference for such results, in general, for parabolic function spaces?



Interpolation and embeddings for parabolic function spaces
No. Take $n=2$ and define $$f(t,x) = \frac{1}{\sqrt{t^2 + \left| x \right|^2}}$$ Then for $1 \leq q < 2$
\begin{align\*}
 \int\_{B\_1(0)} f(t,x)^q dx 
 &= 2\pi \int\_0^1 \frac{r}{(t^2 + r^2)^{q/2}} dr \\
 &= 2\pi \left. \frac{1}{q+2} (t^2 + r^2)^{-q/2 + 1} \right|\_0^1
\end{align\*}


and so $f(t,x) \in L^\infty(0,T;L^q)$


If $p=2$ then 
\begin{align\*}
 \int\_{B\_1(0)} f(t,x)^2
&= 2\pi \left[\log(t^2 + 1) - \log(t^2)\right]
\end{align\*}


Now, $\log(t)$ is integrable on $(0,T)$ and so $f(t,x) \in L^2(Q)$. However, $\log(t)$ is clearly not bounded and so you do $\bf{not}$ get $f(t,x) \in L^\infty(0,T;L^2)$ as you wanted.


