# 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 $L^1$.

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.

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