Prove there exist a unique $u\in H^1$ such that $\int_{\Omega}(\kappa\nabla u\cdot\nabla v + \frac{1}{\kappa}uv) =\ \int_{\Omega}fv$ for $f\in L^2$ I need help with this problem from my homework

Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ and let $\kappa : \Omega\rightarrow\mathbb{R}$ be a continuous function, such that there's constants $M, \beta > 0$, such that $\beta \leq \kappa(x) \leq M$ for each $x\in\Omega$. Show, using some duality result,  that for each $f\in L^2(\Omega)$, there exists a unique $u\in H^1(\Omega)$ such that
$$\int_{\Omega}\left\{\kappa\nabla u\cdot\nabla v + \frac{1}{\kappa}uv\right\}\ =\ \int_{\Omega}fv,\;\;\; \forall\ v\in H^1(\Omega),$$
and
$$\|u\|_{H^1(\Omega)}\ \leq\ \left\{\frac{M}{\min\left\{\beta, \frac{1}{M}\right\}}\right\}^{1/2}\|f\|_{L^2(\Omega)}.$$

Please somebody can help me. Thanks in advance.
 A: This is a community-wiki answer trying to remove this question from the unanswered queue.


Question: Show, using some duality result, that for each $f\in L^2(\Omega)$, there exists a unique $u\in H^1(\Omega)$ such that $u$ is the weak solution to the $\kappa$-weighted elliptic equation, and the solution estimate in $H^1$-norm.

Like you said, this can be proved by Riesz. 
First, we have $f\in L^2(\Omega)\subset (H^{1}(\Omega))'$. This is true because (with slight abuse of notation), we can induce a bounded linear functional on $H^1(\Omega)$ by defining $f(v): = \int_{\Omega} fv$ for any $v\in H^1(\Omega)$.
Second we can show that 
$$
(u,v)_{\kappa}:= \int_{\Omega}\left\{\kappa\nabla u\cdot\nabla v + \frac{1}{\kappa}uv\right\}
$$
is an inner product on $H^1(\Omega)$. This is shown in your other question: How to prove $\int_{\Omega}\frac{1}{\kappa}uv\ +\ \int_{\Omega}\kappa\nabla u\cdot\nabla v$ is an inner product in $H^1$ .
Now applying Riesz representation theorem, there exists a unique $u$ such that 
$$
(u,v)_{\kappa} = f(v) \quad \text{for any }v\in H^1(\Omega). \tag{1}
$$
This gives us the existence of a unique weak solution, which answers your first question.
The estimate is obtained by letting $u=v$ in (1):
$$
M^{-1}\|u\|_{L^2(\Omega)}^2\leq \int_{\Omega}\frac{1}{\kappa} u^2 \leq (u,u)_{\kappa} = \int_{\Omega}fu\leq \|f\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)}
\\
\implies \|u\|_{L^2(\Omega)}\leq M\|f\|_{L^2(\Omega)}.
$$
Yet we have
$$
\min\{\beta,M^{-1}\} \|u\|_{H^1(\Omega)}^2 \leq (u,u)_{\kappa} = \int_{\Omega}fu\leq \|f\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \leq M \|f\|_{L^2(\Omega)}^2.
$$
Therefore:
$$
\|u\|_{H^1(\Omega)}\leq \left\{\frac{M}{\min\left\{\beta, M^{-1}\right\}}\right\}^{1/2}\|f\|_{L^2(\Omega)}.
$$
A: To do the second part, let $v=u$ and just bound below by the bounds on k for each term on the left hand side. Then you take the min of them and the $H^1$ norm basically pops out. Use Cauchy-Schwarz on the right. 
The first part will follow similarly. Use the above strategy to get a coercive bound on the left hand side and use Lax-Milgram. 
