Show that $||u||_{L^{1}(I)} \le \frac{1}{k} ||f||_{L^{1}(I)}$ Suppose there exists a unique $u \in H_0^{1}(I)$ satisfying $$\int_{I}u'v'+k\int_{I}uv=\int_{I}fv, \forall v \in H_0^1(I)$$


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*Show that $$||u||_{L^{1}(I)} \le \frac{1}{k} ||f||_{L^{1}(I)}$$


In the hint it is given : Fix a function $ \gamma \in C^1(\mathbb{R},\mathbb{R})$ such that $\gamma'(t) \ge 0, \forall t \in \mathbb{R}, \gamma(0)=0, \gamma(t)=1 ,\forall t \ge 1, \gamma(t)=-1 ,\forall t \le -1$. Take $v=\gamma(nu)$ and let $n \to \infty$. I am not able to use the hint. 


*Assume now that $f \in L^p(I)$ with $1 \lt p \lt \infty$. Show that there exists a constant $\delta \gt 0$ independent of $k$ and $p$ such that $$||u||_{L^p(I)} \le \frac{1}{k+\frac{\delta}{pp'}}||f||_{L^p(I)}$$ 


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*Do you know the weak derivative of $\gamma(nu)$? Using this, you can show for $V = \gamma(n u)$
$$\int_I u' v' \ge 0$$
and, then, the claim follows by simply $n \to \infty$ (and the dominated convergence theorem).

*What about $v = \gamma(n \, u)^q$ for suitably chosen $q$?
