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Let $d \geq 1$, $\Omega \subset \mathbb{R^d}$ be a bounded domain and let $\phi : [0,T]\times \Omega \mapsto \mathbb{R}$ be a measurable and bounded function. Assume that the following differential inequality holds for all $t \in [0,T]$: $$ \frac{d}{dt}\lVert\phi(t)\rVert_{L^1(\Omega)} \leq K \lVert\phi(t)\rVert_{L^2(\Omega)} $$ where $K > 0$ is a constant.

Is there a useful adaptation of Gronwall's lemma for this type of inequality ?

What I mean by useful is to be able to bound the $L^1$ norm of $\phi(t)$ by some $L^p$ norm of $\phi(0)$, for some $p \in (1,+\infty)$ and up to a multiplicative constant which may depend on $T$. That is, an inequality of the form $$ \lVert\phi(t)\rVert_{L^1(\Omega)} \leq C_T\lVert\phi(0)\rVert_{L^p(\Omega)} $$ for any $t \in [0,T]$.

Any ideas are welcome. The difficulty here lies in the fact that the $L^1$ and $L^2$ norms are not equivalent and the inequality cannot reduce to the classical Gronwall's lemma. I tried writing out the norms and carry the analysis from there, without much success though.

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    $\begingroup$ I made the same comment some days ago that Grönwall changed the spelling of his name to Gronwall when he emigrated from Sweden to the US and he always published under the latter name. Is there any reason not to respect his decision? I know the wikipedia entry but, e.g. the Italian and German pages are about Gronwall's inequality. $\endgroup$
    – Jochen Wengenroth
    Commented May 17 at 7:03
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    $\begingroup$ @JochenWengenroth: I actually didn't know this bit of history; for other's reference, this was discussed in the BAMS memoriam article about him. From now on I may go ahead and just type Gronwall instead; saves me a few keystrokes too. $\endgroup$
    – Willie Wong
    Commented May 17 at 7:47
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    $\begingroup$ @JochenWengenroth My ignorance of that matter was the reason, I've seen it written both ways in various places and always wrongly assumed "Grönwall" was the correct spelling. Thanks for pointing that out, I edited the post. $\endgroup$
    – Theleb
    Commented May 17 at 9:19
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    $\begingroup$ @JochenWengenroth, the change Gronwall -> Grönwall was my responsibility, not the asker's, and I apologise. I always check before making such changes, and figured it was safe to rely on Wikipedia for this, but I did not know the history you mentioned. Thank you for your encouragement to be respectful of people's choice of name (a respect I meant to, but accidentally did not, pay). $\endgroup$
    – LSpice
    Commented May 17 at 11:33

2 Answers 2

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The answer is no. E.g., suppose that $\Omega=(0,1)$, $T=1$, and $$\phi(t,x)=\min(M,t^3x^{-3/4}) =\begin{cases} M&\text{ if }x\le x_t:=t^4/M^{4/3}, \\ t^3x^{-3/4}&\text{ if }x\ge x_t; \end{cases}$$ here and in what follows, $M:=12^3$, $t\in[0,1]$, and $x\in(0,1)$. Then $$\frac d{dt}\,\|\phi(t)\|_1 =\int_{x_t}^1 3t^2 x^{-3/4}\,dx \le \int_0^1 3t^2 x^{-3/4}\,dx=12t^2$$ and $$\|\phi(t)\|_2^2\ge\int_{x_t}^1 t^6 x^{-3/2}\,dx \ge M^{2/3}t^4,$$ so that $$\frac d{dt}\,\|\phi(t)\|_1\le12t^2=M^{1/3}t^2\le\|\phi(t)\|_2.$$ However, $\phi(0)=0$ and $\|\phi(t)\|_1>0$ if $t>0$, so that the inequality $$\|\phi(t)\|_1\le C(t)\|\phi(0)\|_p$$ fails to hold for any $p>0$, any $t\in(0,1]$, and any real $C(t)$.

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  • $\begingroup$ Thanks for your complete answer ! Afterwards, I also thought of a counter-example. Take a $phi(t)$ as the indicator function of a set of size $t^2$ inside $\Omega$, the differential inequality is verified while starting at $0$... $\endgroup$
    – Theleb
    Commented May 17 at 8:01
  • $\begingroup$ @Theleb : I don't understand your comment. However, if you are satisfied with this answer, these guidelines may be relevant now. $\endgroup$
    – Iosif Pinelis
    Commented May 17 at 12:19
  • $\begingroup$ OK. There is a typo : $t^\frac 23$ --> $t^2$. Your answer provides a counter-example where $\phi$ is continuous. I posted another answer which (in my opinion) is less involved computationally (since $\phi$ is an indicator function there). $\endgroup$
    – Theleb
    Commented May 17 at 14:32
  • $\begingroup$ @Theleb : Thank you for your comment. The typo is now fixed. I agree with what you said about your example. $\endgroup$
    – Iosif Pinelis
    Commented May 17 at 19:33
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NO : let's construct a counter-example. Let $\Omega = (0,1)$ and $T = 1$, and for $(t,x) \in [0,T]\times \Omega$, define $\phi(t,x) = 1$ if $x \in [0,t^2]$, and $\phi(t,x) = 0$ otherwise.

For any $t \in [0,T]$, $$ \|\phi(t)\|_1 = t^2 $$ hence $$ \frac{d}{dt}\|\phi(t)\|_1 = 2t $$

Moreover, $$ \|\phi(t)\|_2 = \left(\int_0^{t^2} 1^2 ds\right)^\frac 12= t $$

Now the differential inequality is verified with $K = 2$ (it is in fact an identity), however $\|\phi(0)\|_p = 0$ for all $p \geq 1$ while $\|\phi(t)\|_1 \neq 0$ for all $t \in (0,1]$.

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