prove that there exists a unique $u \in H_0^{1}(I)$ satisfying $\int_{I}u'v'+k\int_{I}uv=\int_{I}fv, \forall v \in H_0^1(I)$

Question

Let $I=(0,1)$ and fix a constant $k \gt 0$.

Given $f \in L^{1}(I)$ prove that there exists a unique $u \in H_0^{1}(I)$ satisfying $$\int_{I}u'v'+k\int_{I}uv=\int_{I}fv, \forall v \in H_0^1(I)$$

For this I defined $$a(u,v)=\int_{I}u'v'+k\int_{I}uv$$ I showed that $a(u,v)$ is a bilinear and coercive along with its continuity. I want to appeal to Lax-Miligram Theorem to settle the claim but alas I am unable to show that $f$ is a continuous linear functional on $H_0^1(I)$.

Answer 1

Keep in mind that there is a constant $C$ with the property that $\|v\|_\infty \le C \|v\|_{H^1}$ for all $v \in H^1(I)$. If $f \in L^1(I)$ then $$\left| \int_I fv \right| \le \|f\|_1\|v\|_\infty \le C \|f\|_1 \|v\|_{H^1}.$$