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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?

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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.

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  • $\begingroup$ There's a typo: the conclusion for the first part $1\leq q<2$ should read: $f\in L^{\infty}(0,T;L^q)$ for all $1\leq q$, not $f\in L^2(0,T;L^2)$. $\endgroup$ Jun 2, 2014 at 17:55
  • $\begingroup$ Thanks, I tried something similar, but for some reason did not think of this particular function. $\endgroup$ Jun 2, 2014 at 18:21
  • $\begingroup$ @ k3thomps: sure, nice counter-example! By the way: when thinking of this problem I tried to prove the following, which should be obvious (at least to me). If we further assume $L^{\infty}(0,T;L^q)$ bounds on $f$ uniformly in $q<2$ then we should have $f\in L^{\infty}(0,T;L^2)$ as well (just by analogy with the 'stationary' classical $|f|_{L^p}=\lim _{q\to p}|f|_{L^p}$). However, when I tried to write this down rigorously I quickly ran into quantifier problems (like: $\exists E\subset(0,T)$ of full measure s.r $\forall q<2$..., or $\forall q<2\,\exists$ a full measure set $E_q$ etc). Any idea? $\endgroup$ Jun 2, 2014 at 18:42
  • $\begingroup$ Ooops, I meant of course $|f|_{L^p}=\lim_{q\to p}|f|_{L^q}$. Sorry about that. $\endgroup$ Jun 2, 2014 at 18:56
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    $\begingroup$ @leomonsaingeon How about showing that by duality, ie. using the property that $f\in L^\infty L^2 \Longleftrightarrow \int fg \leq C\|g\|_{ L^1L^2} $ for all smooth $g$? For smooth $g$ it's straightforward to show the convergence of the $L^1L^{q'}$-norm to the $L^1L^2$-norm. $\endgroup$ Jun 4, 2014 at 20:58

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