Exercise from Brezis:
Let $I=(0,1)$ and fix a constant $k>0$. Given $f\in L^{1}(I)$ prove that there exists a unique $u\in H_{0}^{1}(I)$ satisfying \begin{align} \int_{I}u'v'+k\int_{I}uv=\int_{I}fv\quad\forall v\in H_{0}^{1}(I). \end{align}
I have shown that the bilinear form, \begin{align} a(u,v)=\int_{I}u'v'+k\int_{I}uv, \end{align} is continuous, coercive and symmetric. However, I do not know how to deal with $f\in L^{1}(I)$. How can I find a functional $\varphi\in (H_{0}^{1}(I))^{*}$ such that there is a unique $f\in L^{2}(I)\subset H^{-1}(I)$ giving the integral representation, \begin{align} \langle\varphi,v\rangle=\int_{I}fv\quad\forall v\in H^{1}_{0}(I)? \end{align}
First, you define $$\phi(v):= \int_I fv\ dx.$$ Second, you have to verify that $\phi$ is defined on $H^1_0(I)$, linear, and bounded. Linearity is obvious. For well-definedness and boundedness, you might check for suitable embedding theorems (and H\"older inequality).