Why do you want to avoid time discretization?

Actually, @piero-dancona says it much nicer as an answer to your other question at mathoverflow.net/a/159500/16530.

Why is uniqueness for $-\Delta \Psi = u$ with $u\in L^1(R^d)$ an issue? If I have two solutions $\Psi_1$ and $\Psi_2$ of the equation, both in the sense of distributions (so $\Psi_{1,2}$ only need be $L^1_{\mathrm{loc}}$), then the difference $\Psi=\Psi_1-\Psi_2$ satisfies $\Delta\Psi=0$ in the sense of distributions. Then $\Psi$ is $C^\infty$, as can be seen by regularizing $\Psi$ by convolution. If $\Psi$ is bounded and harmonic, then it is constant ... Maybe the boundedness of $\Psi$ is the issue?

@leomonsaingeon How about showing that by duality, ie. using the property that $f\in L^\infty L^2 \Longleftrightarrow \int fg \leq C\|g\|_{ L^1L^2} $ for all smooth $g$? For smooth $g$ it's straightforward to show the convergence of the $L^1L^{q'}$-norm to the $L^1L^2$-norm.

You're right, some control on time variation is necessary. But what is your question?}

Could the difference be whether the gradient flow is with respect to the $L^2(g_0)$ metric or the $L^2(g)$ metric?

@Kofi, you'll need some kind of condition to prevent e.g. convergence of convolution kernels to a delta function: if $p_n(x) = n p(nx)$, with $p\in C_c^\infty(\mathbb R)$, then convolution with $p_n$ satisfies all your assumptions, with the exception (I assume) of the limit operator being 'smoothing'. That suggests that you need to quantify what 'smoothing' means, and how the 'smoothing' nature of $p_n$ is conserved in the limit. Can you say something about that?

I deleted my answer because too many misunderstandings went into it.

One classical reference for this is the book by Jacques-Louis Lions, Quelques methodes de resolution de problèmes aux limites non linéaires.

You need to make your question more explicit. What exactly are you looking for?

Thanks @Theo and @Gerard - I obviously overlooked Theorem IV.6.2, possibly because of the different notation. Thanks for pointing that out!

Of course, in higher dimensions the gradient can not be just any vector-valued $L^\infty$-function $f = (f_i)$, since it satisfies the distributional identity $ \dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}. $ So you don't get the whole of $L^\infty$, only those functions that satisfy this identity.

Tapio, why do you want the entropy of your finitely additive translation-invariant measure to be finite? In the limit as $n\to\infty$ the Shannon entropy of the uniform measure on $[n]$ converges to infinity. Wouldn't you therefore expect that such a `uniform' measure on $\mathbb N$ should have infinite entropy?