Assume that $ \Omega $ is a smooth bounded domain in $ \mathbb{R}^n $. Consider a functional
$$
\mathcal{F}(u)=\int\_\Omega(|\nabla u|^2+h^{-1}|u-u\_0|^2) \, dx
$$
where $ h>0 $ is a parameter and $ u\_0\in H\_0^1(\Omega) $. For the minimizing problem
$$ 
\min\_{u\in H\_0^1(\Omega)}\mathcal{F}(u),
$$
it is easy to get by standard arguments that there is a unique minimizer $ u\_h\in H\_0^1(\Omega) $, i.e. $ u\_h=\operatorname{argmin}\_{u\in H\_0^1(\Omega)} \mathcal{F}(u) $. A natural observation is that $ \|u\_h-u\_0\|\_{L^2(\Omega)}\to 0 $. I want to ask the following question.



> 
> Let $ h\_0>0 $. Assume that $ h\to h\_0 $. If there is the result $ \|u\_{h}-u\_{h\_0}\|\_{L^2(\Omega)}\to 0 $?
> 
> 
> 


Such question is natural and intuitively true. However I cannot solve it, can you give me some hints or references?



Limit of minimizers of a class of functionals
By simple calculations, we can obtain that $ u\_{h} $ and $ u\_{h\_0} $ satisfy the following Euler-Lagrange equation
\begin{align}
\frac{u\_h-u\_0}{h}-\Delta u\_{h}=0,\\
\frac{u\_{h\_0}-u\_0}{h}-\Delta u\_{h\_0}=0.
\end{align}
Then it can be got that
$$
\begin{aligned}
0&=\frac{u\_h-u\_0}{h}-\frac{u\_{h\_0}-u\_0}{h\_0}-\Delta(u\_{h}-u\_{h\_0})\\
&=\frac{u\_h-u\_0}{h}-\frac{u\_h-u\_0}{h\_0}+\frac{u\_h-u\_0}{h\_0}-\frac{u\_{h\_0}-u\_0}{h\_0}-\Delta(u\_{h}-u\_{h\_0})\\
&=(u\_h-u\_0)\frac{h\_0-h}{hh\_0}+\frac{u\_h-u\_{h\_0}}{h\_0}-\Delta(u\_h-u\_{h\_0}).
\end{aligned}
$$
Testing the above equation by the function $ u\_h-u\_{h\_0} $, we obtain that
$$
\int\_{\Omega}|\nabla(u\_h-u\_{h\_0})|^2dx+\int\_{\Omega}\frac{(u\_{h}-u\_{h\_0})^2}{h\_0}dx\leq\int\_{\Omega}\frac{h-h\_0}{hh\_0}(u\_h-u\_0)(u\_h-u\_{h\_0})dx.
$$
By Cauchy inequality, we have
$$
\frac{1}{2}\int\_{\Omega}(u\_h-u\_{h\_0})^2dx\leq\frac{(h-h\_0)^2}{h^2}\int\_{\Omega}(u\_h-u\_0)^2dx.
$$
From this, we can complete the proof.


