If $u \in L^2(0,T;L^2)$ and $u' \in L^2(0,T;H^{-1})$, is $u \in C^0([0,T];V)$ for some space $V$?

Question

If $u \in L^2(0,T;L^2)$ has weak derivative $u' \in L^2(0,T;H^{-1})$, is $u \in C^0([0,T];V)$ for some Banach space $V$? For what $V$ Lebesgue spaces does this hold? I cannot find any results.

Answer 1

Generally, for $\,V\,$ to be a Lebesgue space, it must be $\,u\in L^2(0,T;H^1)\,$ instead of $\,u\in L^2(0,T;L^2)$, in which case $\,V=L^2$. Otherwise, say, when $\,u\in L^2(0,T;H^s)\,$ instead of $u\in L^2(0,T;L^2)$ with some $s\in (0,1)$, the choice of $V$ cannot be tetter than $V=H^{-\sigma}$ with $\sigma=(1-s)/2$. In your particular case of the embedding $L^2(0,T;L^2)\cap H^1(0,T;H^{-1})\hookrightarrow C([0,T];V)$, to show that $V=H^{-1/2}$, assume that domain $\Omega=\mathbb{R}^n$, so that elements of $L^2$ and $H^1$ are defined on $\mathbb{R}^n$. By $\,\Lambda\,$ denote a pseudodifferential operator defined as $\,F[\Lambda u]=F^{-1}\bigl[\sqrt{1+|\xi|^2}\widehat{u}(t,\xi)\bigr]$, where $\,\widehat{u}=F[u]\,$ denotes the Fourier transform on $\,\mathbb{R}^n\,$ with its inverse $\,F^{-1}$. Observe that $$ f\overset{\rm def}{=}u_t + \Lambda\,u\in L^2(0,T;H^{-1}). \tag{1} $$ It is not difficult to construct a sequence $\,\{f_n\}\,$ of the elements $\,f_n\in C([0,T];H^{-1})\,$ such that $\,f_n\to f\,$ in $\,L^2(0,T;H^{-1})$. Obviously, for each index $n$, there exists a unique solution $\,u_n\in C^1([0,T];H^{-1})\,$ of the Cauchy problem $$ \begin{cases} \frac{\partial u_n}{\partial t}+\Lambda\,u_n=f_n\,,\quad t\in (0,T),\\ u_n|_{t=0}=0\,, \end{cases} $$ where without loss of generalty the intial data are assumed homogeneous. Note that the case of inhomogeneous initial data readily reduces to homogeneous by employing an even extension of $\,u\,$ from $\,(0,T)\,$ to $\,(-T,T)\,$ w.r.t. variable $\,t\,$ followed by multiplying the extended $\,u\,$ by some cut-off function $\,\eta\in C^{\infty} [-T,T]\,$ such that $\,\eta=1\,$ on $\,[0,T]\,$ while $\,\eta=0$, say, on $\,[-T,-T/2]$.

After applying Fourier transform, the Cauchy problem for $u_n$ takes the form $$ \begin{cases} \frac{\partial \widehat{u}_n}{\partial t}+\sqrt{1+|\xi|^2}\widehat{u}_n=\widehat{f}_n\,,\quad t\in (0,T),\\ \widehat{u}_n|_{t=0}=0, \end{cases} $$ whence readily follows $$ \begin{align} \frac{1}{2}\frac{d\,}{dt}\int\limits_{\mathbb{R}^n}\frac{|\widehat{u}_n(t,\xi)|^2}{\sqrt{1+|\xi|^2}}d\xi+\int\limits_{\mathbb{R}^n}|\widehat{u}_n(t,\xi)|^2\,d\xi\leqslant\int\limits_{\mathbb{R}^n}\frac{|\widehat{f}_n(t,\xi)|\cdot |\widehat{u}_n(t,\xi)|}{\sqrt{1+|\xi|^2}}d\xi\\ \leqslant\frac{1}{2}\int\limits_{\mathbb{R}^n} \frac{|\widehat{f}_n(t,\xi)|^2}{1+|\xi|^2}d\xi + \frac{1}{2}\int\limits_{\mathbb{R}^n}|\widehat{u}_n(t,\xi)|^2 d\xi, \quad\forall\,t\in (0,T), \end{align} $$ which yields the inequality $$ \frac{d\,}{dt}\|u_n\|^2_{H^{-1/2}}+\|u_n\|^2_{L^2}\leqslant \|f_n\|^2_{H^{-1}} \quad\forall\,t\in (0,T). $$ Integrating the latter inequality results in the estimate $$ \max_{t\in [0,T]}\|u_n\|^2_{H^{-1/2}}+\int\limits_0^T\|u_n\|^2_{L^2}\,dt \leqslant\int\limits_0^T\|f_n\|^2_{H^{-1}}\,dt \quad\forall\, n.\tag{2} $$ In the same manner follows the estimate $$ \max_{t\in [0,T]}\|u_m-u_n\|^2_{H^{-1/2}}+\int\limits_0^T\|u_m-u_n\|^2_{L^2}\,dt \leqslant\int\limits_0^T\|f_m-f_n\|^2_{H^{-1}}\,dt \quad\forall\, m,n, $$ which implies that $\,\{u_n\}\,$ is a Cauchy sequence in $\,C([0,T];H^{-1/2})\cap L^2(0,T;L^2)\,$ since $\,\{f_n\}\,$ is a Cauchy sequence in $\,L^2(0,T;H^{-1})\,$. Hence $\,\{u_n\}\,$ does converge to some element of a Banach space $\,C([0,T];H^{-1/2})\cap L^2(0,T;L^2)\,$ that must coincide with the original $u$ due to the fact that the Cauchy problem $$ \begin{cases} w_t+\Lambda\,w=0,\quad t\in (0,T),\\ w|_{t=0}=0 \end{cases} $$ has no weak solution $w\in L^2(0,T;L^2)$ other than the trivial $w=0$, i.e, the integral identity $$ -\int\limits_0^T\int\limits_{\mathbb{R}^n} w\,\varphi_t\,dxdt+\int\limits_0^T \int\limits_{\mathbb{R}^n}w\,\Lambda\varphi\,dxdt=0\quad\forall\,\varphi\in C^{\infty}\bigl([0,T];\mathcal{S}(\mathbb{R}^n)\bigr)\colon\;\varphi|_{t=T}=0 $$ immediately implies that $\,w=0\,$ a.e on $(0,T)\times\mathbb{R}^n$, where $\mathcal{S}(\mathbb{R}^n)$ stands for the Schwartz space of rapidly decreasing functions. Thus follows the embedding $$ L^2(0,T;L^2)\cap H^1(0,T;H^{-1})\hookrightarrow C([0,T];V) $$ in conjunction with the estimate $$ \max_{t\in [0,T]}\|u\|^2_{H^{-1/2}}\leqslant \|f\|^2_{L^2(0,T;H^{-1})} \leqslant C\Bigl(\|u\|^2_{H^1(0,T;H^{-1})}+\|u\|^2_{L^2(0,T;L^2)}\Bigr) $$ that readily follows by $(1)$ and $(2)$.   Q.E.D.