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.)

  • $\begingroup$ 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. $\endgroup$ – C_Al Feb 15 '15 at 13:44
  • $\begingroup$ Also, you write "The last term is equal to..", for this you have to assume that $\nabla f(e)$ exists, right? $\endgroup$ – C_Al Feb 15 '15 at 14:59
  • $\begingroup$ 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. $\endgroup$ – Hachino Feb 15 '15 at 15:27
  • $\begingroup$ 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 $1 < p < q <\infty$. $\endgroup$ – Hachino Feb 15 '15 at 15:29
  • $\begingroup$ @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 $n$. $\endgroup$ – PhoemueX Feb 15 '15 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.