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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?

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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.)

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  • $\begingroup$ 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

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