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To simplify, let us work on $Q_T:=[0,T]\times\mathbb{T}^N$ where $\mathbb{T}^N$ is the $N$-th dimensionnal torus.

Consider $(S_n)_n$ a sequence of $L^1(Q_T)$ and $(z_n)_n$ the sequence of solutions to the heat equation $\partial_t z_n - \Delta z_n = S_n$ (with $0$ as initial data).

Is it true that if $(S_n)_n$ is bounded in $L^1(Q_T)$, then $(z_n)_n$ converges (up to a subsequence) a.e. ? If it is not sufficient, does $(S_n)_n$ weakly converging in $L^1(Q_T)$ sufficient ?

I expect some compactness for the operator $S\mapsto z$ but I did not manage to find some reference for the $L^1(Q_T)$ case.

Thanks for any help !

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  • $\begingroup$ Apparently $S \mapsto z$ is compact by Corollary 5.1 in [S. P. Eveson Proceedings of the American Mathematical Society Vol. 123, No. 12 (Dec., 1995), pp. 3709-3716], jstor.org/stable/2161898 $\endgroup$ Aug 11, 2017 at 17:56
  • $\begingroup$ @ Mateusz: I don't see how this covers the heat equation? It seems to me that Eveson's setting - i.e. kernels $k(x,y)$ with $(x,y)\in\Omega\times\Omega$ in a domain $\Omega\subset\mathbb R^N$ - rather corresponds to Green's functions for elliptic problems. Am I missing something? $\endgroup$ May 15, 2019 at 1:17
  • $\begingroup$ If this question ever finds a definite answer: can we extend to measure right-hand sides (with total variation in place of $L^1$ norm)? $\endgroup$ May 15, 2019 at 1:19

1 Answer 1

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Yes the input-output operator $S \mapsto z$ is compact with the $L^1$-controls. For the proof see for instance Theorem 1.2 in

R. Triggiani, A note on the lack of exact controllability for mild solutions in Banach spaces, SIAM J. Control Optimization, 15 (1977).

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