Convergence in the weak-star sense of measures and $ \int_{\Omega} \sqrt{1+u_k^2} \to \int_{\Omega} \sqrt {1+u_k^2}$ gives convergence in $L^1$.

Question

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.

Answer 1

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