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Let $k \in {L^\infty }\left( {0,T} \right)$ and we assume that $$\phi :t \mapsto u\left( t \right) + \int_0^t {k\left( s \right)u\left( s \right)ds} \in {L^\infty }\left( {0,T;{L^2}\left( \Omega \right)} \right).$$ Can we prove that $u \in {L^\infty }\left( {0,T;{L^2}\left( \Omega \right)} \right)$.

My approach is here. We have $$\left\| {u\left( t \right)} \right\| = \left\| {\phi \left( t \right) + \int_0^t {k\left( s \right)u\left( s \right)ds} } \right\| \leqslant \left\| {\phi \left( t \right)} \right\| + \int_0^t {\left| {k\left( s \right)} \right|\left\| {u\left( s \right)} \right\|ds} \leqslant {\left\| \phi \right\|_{{L^\infty }\left( {0,T;{L^2}} \right)}} + {\left\| k \right\|_{{L^\infty }\left( {0,T} \right)}}\int_0^t {\left\| {u\left( s \right)} \right\|ds} .$$ Use Gronwall inequality, we get $\left\| {u\left( t \right)} \right\| \leqslant {\left\| \phi \right\|_{{L^\infty }\left( {0,T;{L^2}} \right)}}\exp \left( {{{\left\| k \right\|}_{{L^\infty }\left( {0,T} \right)}}T} \right)$. Then we have $u \in {L^\infty }\left( {0,T;{L^2}} \right)$.

But i cant prove $u\left( t \right) \in {L^2}$.

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    $\begingroup$ Doesn't $\lVert u(t)\rVert \le \lVert\phi\rVert_\infty\exp(\lVert k\rVert_\infty T)$ imply in particular that $\lVert u(t)\rVert < \infty$, hence that $u(t) \in \operatorname L^2$? Or is $\lVert u(t)\rVert$ not the $\operatorname L^2$ norm? $\endgroup$
    – LSpice
    Commented Dec 18, 2019 at 18:03
  • $\begingroup$ At first we dont have $u\left( t \right) \in {L^2}\left( \Omega \right)$. So we cant define $\left\| {u\left( t \right)} \right\|$ and we dont have $$\left\| {u\left( t \right)} \right\| \leqslant {\left\| \phi \right\|_{{L^\infty }\left( {0,T;{L^2}} \right)}} + {\left\| k \right\|_{{L^\infty }\left( {0,T} \right)}}\int_0^t {\left\| {u\left( s \right)} \right\|ds} .$$ $\endgroup$
    – Trần Quang Minh
    Commented Dec 19, 2019 at 1:06
  • $\begingroup$ Which space does $u(t)$ belong to, to start wth? When you say that $\Phi\in L^\infty L^2$ you define the Bochner integral $\int_0^t ku$, so implicitly you must assume that $u$ belongs to some Bochner space in order to ascribe a menaing this integral. $\endgroup$
    – leo monsaingeon
    Commented May 18, 2020 at 12:54
  • $\begingroup$ I strongly suspect that you should first assume that $u\in L^1_{loc}L^2$ in order to make the question meaningful in the first place (in particular $u(t)$ is unambiguously defined as an $L^2(\Omega)$ for a.e. time), and only then try to upgrade the time regularity to $L^\infty$ $\endgroup$
    – leo monsaingeon
    Commented May 18, 2020 at 13:01
  • $\begingroup$ For example, you can check by hand that, for any arbitrary point $x\in\Omega$, the distribution-valued function $u(t):=\exp(-\int _0^1 f)\delta_{x}$ solves $u'+ku=0$, hence $\phi(t)=u(t)+\int_0^t k u\equiv 0$ has all the regularity that one may think about. However of course this specific function $u$ does not belong to $L^\infty L^2$. $\endgroup$
    – leo monsaingeon
    Commented May 18, 2020 at 13:06

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