7
$\begingroup$

I have a question, which is exercise 1.20 of the following book:

Functions of Bounded Variation and Free Discontinuity Problems.

Let assume $\Omega$ is a bounded subset of $\mathbb{R}^n$ and $u_k, u \in L^1(\Omega)$ and $u_k$ converge to $u$ in the weak star sense of measures as follow: $$ \int_{\Omega} u_k \phi \to \int_{\Omega} u \phi \qquad \forall \phi \in C_c^{\infty}(\Omega).$$ Also assume that $$ \int_{\Omega} \sqrt{1+u_k^2} \to \int_{\Omega} \sqrt{1+u^2}.$$ Then we want to show that we have strong convergence in $L^1(\Omega)$ too.

There is a hint which says first show that $$\sqrt{1+u_k^2}+\sqrt{1+u^2}-2\sqrt{1+\left(\frac{u+u_k}{2}\right)^2} \to 0 $$ in $L^1(\Omega)$. This is okay but I don't know how to use this last one to conclude the result.

$\endgroup$
2
+50
$\begingroup$

Let $g(x)=\sqrt{1+x^2}$ and $g_2(x,y)=g(x)+g(y)-2g((x+y)/2).$

(Here is a summary of the following argument. The function $g$ is convex, and strongly convex on compact sets, so if $g_2(x,y)\to 0$ and $x$ is bounded then $|x-y|\to 0.$ You know that $g_2(u,u_k)$ tends to zero in $L^1,$ so it tends to zero in measure, so $u_k\to u$ in measure. This combined with $\int g(u_k)\to \int g(u)$ gives $u_k\to u$ in $L^1.$)

Fix $\epsilon>0.$ Pick $N$ such that $|u|\leq N$ except on a set $E_1$ of measure $\leq\epsilon/2.$ Use $$g_2(x,y)=\int_{\min(x,y)}^{\max(x,y)} g''(t)\min(|t-x|,|t-y|)dt\geq \int_{\min(x,(x+y)/2)}^{\max(x,(x+y)/2)} g''(t)|t-x|dt$$

to get a $\delta=\delta(\epsilon,N)$ such that $|x|\leq N$ and $g_2(x,y)\leq\delta$ implies $|x-y|\leq \epsilon/2.$ Pick $k$ sufficiently large such that $g_2(u,u_k)\leq\delta$ except on a set $E_2$ (depending on $k$) of measure $\leq\epsilon/2.$ Then $|u-u_k|\leq\epsilon$ except on $E:=E_1\cup E_2,$ which has measure $\leq\epsilon.$

The $L^1$ norm of $u$ on $E$ (or any set of measure $\epsilon$) tends to zero as $\epsilon\to 0.$ The $L^1$ norm of $u_k$ on $E$ also tends to zero because

$$\int_E|u_k|\leq\int_E g(u_k)=\int_{\Omega} g(u_k)-\int_{\Omega\setminus E}g(u_k)\approx \int_{\Omega} g(u)-\int_{\Omega\setminus E}g(u)\approx 0.$$

So $\|u_k-u\|_1\to 0.$

$\endgroup$

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.