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?


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.

  • $\begingroup$ 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 $f\in L^2(0,T;L^2)$. $\endgroup$ – leo monsaingeon Jun 2 '14 at 17:55
  • $\begingroup$ Thanks, I tried something similar, but for some reason did not think of this particular function. $\endgroup$ – Juhana Siljander Jun 2 '14 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? $\endgroup$ – leo monsaingeon Jun 2 '14 at 18:42
  • $\begingroup$ Ooops, I meant of course $|f|_{L^p}=\lim_{q\to p}|f|_{L^q}$. Sorry about that. $\endgroup$ – leo monsaingeon Jun 2 '14 at 18:56
  • 2
    $\begingroup$ @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. $\endgroup$ – Mark Peletier Jun 4 '14 at 20:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.