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Let $T>0$ and $f\in C(\mathbb R, L^{2}(\mathbb R))$ with the following property:

Put $g(t):= \sup\limits_{0\leq \tau\leq t} \|f(\tau)\|_{X},$ where $X \subset L^{2}$ and $X$ is a Banach Space.

For $0\leq t \leq T, $ $0<p<\infty,$ $g(t)^p \leq 1+ \int_{0}^{t} g(\tau)^{p} d\tau.$

My Question is: Can we expect to show that $g\in L^{\infty}([0, T])$?

My Vague Idea: I think, we need to use Gronwall Lemma for the integral form, but I am little confuse which form I should consider, and how to apply.

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    $\begingroup$ Well, simply the fact that $g(t)^p\leqslant C$ for every $t$ in $[0,T]$, with $C=1+\int_0^Tg(\tau)^pd\tau$, allows to conclude, no? $\endgroup$ – Did Mar 26 '16 at 10:50
  • $\begingroup$ @Did: Why $C< \infty$? $\endgroup$ – Inquisitive Mar 26 '16 at 10:55
  • $\begingroup$ Depends on your hypotheses, that $g(T)$ is finite suffices (anyway, Gronwall does not seem needed). In the other direction, note that $g(0)=0$, $g(\tau)=+\infty$ for every $\tau>0$ does solve the integral inequality... $\endgroup$ – Did Mar 26 '16 at 10:59
  • $\begingroup$ @Did: Thanks: But to apply Gronwall do we need continuity assumption on the hypothesis? That is my confusion. $\endgroup$ – Inquisitive Mar 26 '16 at 11:02
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    $\begingroup$ I agree that the paper lacks justification in the application of Gronwall's lemma. What has been proved there is the implication $$\omega \in L^{\alpha'}_{\rm loc}([0, \infty))\ \Rightarrow \omega \in L^\infty_{\rm loc}([0, \infty)).$$ So either the paper is wrong or (more probably) the $L^{\alpha'}$ integrability of $\omega$ is obvious. $\endgroup$ – Giuseppe Negro Mar 26 '16 at 16:04

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