# Identifying the weak limit of a gradient (Bochner spaces)

Let $X=L^2(0,T;L^2(\Omega))$ for an unbounded domain $\Omega$. Let $f_n, f:\mathbb{R} \to \mathbb{R}$ be functions with $f_n \to f$, $f_n(0)=f(0)=0$ and $f_n$ Lipschitz with Lipschitz constant depending on $n$. In fact $f_n(x) := \int_0^x |T_n((|s|-\frac 1n)^+ + \frac 1n)|^{-\frac{1}{2}}$ where $T_n(x) = x$ for $|x| \leq n$ and $T_n(x) = n$ otherwise (the usual truncation function).

I have the following convergence results: $$e_n \to e \quad\text{in X}$$ $$\nabla e_n \rightharpoonup \nabla e\quad\text{in X}$$ $$\nabla f_n(e_n) \rightharpoonup f^*\quad\text{in X}$$ $$f_n(e_n) \to f(e) \quad\text{pointwise a.e.}$$ I wish to idenfify $f^*$ with $\nabla f(e)$.

I also have additional uniform bounds on $f_n(e_n)$ and $e_n$ in the space $L^\infty(0,T;L^\infty(\Omega))$. Unfortunately since the domain is unbounded we can't say anything about $f_n(e_n)$ being bounded in $L^2$.

A DCT argument doesn't work either.

If it helps,

Does anyone have any ideas or techniques to do this?

Your uniform bounds in $L^{\infty}_t L^{\infty}_x$ will be of great help here. First, let us choose some big radius $R > 0$ and restrict our attention to the ball $B(0,R)$ instead of $\Omega$.

UPDATE : Here is a second attempt of a proof, with the same idea as before.

Let $\varphi$ be a function in $\mathcal{D}(]0,T[ \times \Omega)$ and choose $R$ big enough so as to cover the spatial support of $\varphi$.

From the weak convergence in $L^2_t L^2_x$, we know that

$$< \nabla f_n(e_n), \varphi >_{\mathcal{D}', \mathcal{D}} = \int \nabla f_n(e_n) \varphi \to \int f^* \varphi .$$

On the other hand, because you have a uniform bound for $f_n(e_n)$ in $L^{\infty}_t L^{\infty}_x$ and that constants are integrable on $]0,T[ \times B(0,R)$, DCT tells you that

$$\int f_n(e_n) \nabla \varphi \to \int f(e) \nabla \varphi .$$

The last term is equal to $$- < \nabla f(e), \varphi >_{\mathcal{D}', \mathcal{D}}$$

and we conclude that $f^*$ and $\nabla f(e)$ agree as distributions. As they are both functions, they also agree as functions, in $L^2_t L^2_x$ for instance.

Sorry again for the failed attempt, hope this one will be clearer.

(Notice one thing : you only need uniform bounds on $f_n(e_n)$ locally in space and time, not globally.)

• Thank you for the answer! Are you sure that $L^\infty(0,T;L^\infty)$ is compactly embedded in $L^2(0,T;L^2)$ (on a bounded domain)? I couldn't find any references for this. – C_Al Feb 15 '15 at 13:44
• Also, you write "The last term is equal to..", for this you have to assume that $\nabla f(e)$ exists, right? – C_Al Feb 15 '15 at 14:59
• Nope, I'm taking gradients in the space of distributions, thus they always exist. What I could not do, however, is to write an identity like $\nabla f(e) = \nabla e f'(e)$, since the RHS needs some regularity to be well defined. – Hachino Feb 15 '15 at 15:27
• Regarding the compact embedding, that's a rather straightforward argument if you know the compactness criterion in $L^p$ spaces, see here for instance. More generally, on domains of finite measure, not only are $L^p$ spaces decreasing with $p$, but compactness holds at each step, that is, $L^{\infty} \Subset L^q \Subset L^p \Subset L^1$ for $1 < p < q <\infty$. – Hachino Feb 15 '15 at 15:29
• @Hachino: The inclusion is in most cases not compact. Consider e.g. $\Omega = (0,1)$ and $f_n (x)= e^{2 \pi i n x}$. Then $(f_n)_n$ is a bounded sequence in each $L^p(\Omega)$ space, but has no weakly convergent subsequences. To see this, note that $f_n \to 0$ weakly, but $\Vert f_n \Vert_p = 1$ for all $n$. – PhoemueX Feb 15 '15 at 18:06

Allow me to jump in here. Athough I'm usually no big fan of abstract "Bourbaki-style" mathematics myself (I'm more of a hands-on applied mathematician, but that's only my personal taste), I do believe that sometimes abstraction helps getting the big picture and identifying the key steps, rather than rushing into tedious integrations by parts.

This specific post is such a typical situation in PDEs that I believe it is worth dissecting the structure of the problem:

1. we have that some derivative of some quantity converges to some limit, say $$v=\lim \nabla u_n$$, but in a weak topology (Here $$u_n=f_n(e_n)$$, but it doesn't really matter that $$u_n$$ is a nonlinear function of $$e_n$$)
2. we know that the said quantity converges to some limit (here $$u_n=f_n(e_n)\to f(e):=u$$), but in some stronger sense
3. we have a bunch of additional estimates
4. we wish to "pass derivatives to the limit" $$\lim(\nabla u_n)=\nabla(\lim u_n)$$, in other words $$v=\nabla u$$

The case I want to make here is that using a very weak topology often allows to conclude. Indeed, strong results are often available in weaker topologies. (The Banach-Alaoglu theorem is a striking example: bounded sets are of course NOT relatively compact for strong topologies, but they become compact again if you're willing to weaken the topology to weak-*)

Let me illustrate this in the OP's specific framework. An abstract notion appearing implicitly in @Hachino's answer is that of distributions, $$\mathcal D'((0,T)\times\Omega)$$. This is the pivot space, in which everyting actually builds up wonderfully. Note of course that the natural topology of $$\mathcal D'$$ is extremely weak, that's the whole point of L. Schwartz's theory!

There are two deep results in the theory of distributions that I'm going to use here:

1. differentiation is a continuous operation from $$\mathcal D'$$ to itself, namely if $$T_n$$ is a sequence of distributions converging in $$\mathcal D'$$ to a limit $$T$$ then automatically $$\partial_\alpha T_n$$ converges to $$\partial_\alpha T$$ for any multiindex $$\alpha$$. (Since the $$\mathcal D'$$ topology is vey weak it is easier for the natural operations to be continuous)
2. $$L^1_{loc}$$ is continuously embedded into $$\mathcal D'$$ (which also reflects the fact that the topology of $$\mathcal D'$$ is "weak enough")

In the precise setting of the OP, the argument works then as follows:

1. We know that $$\nabla f_n(e_n)\rightharpoonup f^*$$ weakly in $$X$$, hence in particular in the sense of distributions (the topology on $$\mathcal D'$$ is weaker than the weak topology on $$X$$)
2. the pointwise a.e convergence $$f_n(e_n)\to f(e)$$ together with uniform $$L^\infty$$ bounds gives strong convergence in $$L^1_{loc}$$ (this is the Dominated Convergence Theorem, which works fine on bounded domains) hence in the sense of distributions (which, again, is weaker than $$L^1$$. I hope you start seeing my point about weaker topologies, now!)
3. by the two abstract distributional properties that I mentioned earlier we are done, and indeed $$f^*=f(e)$$ in the sense of distributions.

That's it, I hope this (probably overpedantic) answer helps out someone. At least I wish someone explained this to me when I first learnt PDEs. And now I realize how important the theory of distributions is for applied maths, nobody ever told me that back then (and don't get me wrong, the professor was great, probably the greatest I ever had, I just think that we mathematicians should teach more about "metamathematics" in our maths classes)