# For a bounded linear functional $l(v) := (f,v)_\Omega$, do we have $\|l\| \le \|f\|$?

Suppose we have a Poisson equation $$-\Delta u = f$$ on $$\Omega$$ and we want to derive its weak formulation, so we multiply it by an arbitrary test function $$\forall v \in H^1_0$$ and then take integral over $$\Omega$$ \begin{aligned} (-\Delta u,v)_\Omega = (f,v)_\Omega \end{aligned} where we denote by $$(\cdot,\cdot)_\Omega$$ the inner product, and we want to find the weak solution $$u \in H^1_0$$ such that \begin{aligned} a(u,v) = l(v) \end{aligned} where the bilinear form $$a(u,v) := (-\Delta u,v)_\Omega$$ and the linear form $$l(v) := (f,v)_\Omega$$.

Let's focus on the right-hand side. By Cauchy-Schwarz inequality we have $$(f,v)_\Omega \le \|f\|\|v\|$$, while by the boundedness of the linear functional we have $$l(v) \le \|l\|\|v\|$$, where$$\|l\|:= \sup_{\|v\|=1} |l(v)|$$ denotes the operator norm. Note that we also have $$(f,v)_\Omega=l(v)$$. We may thus conclude that $$\|l\| \le \|f\|$$. Is that correct? Can we further obtain $$\|l\|=\|f\|$$?

• $\|\ell\|\leq \|f\|$ is indeed obvious (and you justified it quite correctly). But be careful nevertheless : $a\leq |b|$ doesn't implies that $|a|\leq |b|$. Be here $\ell$ is linear so you can manage the lower bound). For the converse inequality (that is also true), If $E$ is a banach space, then one can prove that $\|x\|_E=\sup_{\|f\|_{E'}\leq 1}|f(x)|,$ where $E'$ is the topological dual of $E$. – Surb Nov 26 '18 at 14:58
• By definition, $l(v) := (f,v)_\Omega$. The estimate $|l(v)|=|(f,v)_\Omega| \le \|f\|\|v\|$ proves two things: First, $l$ is bounded. Second, $\|l\|\leq \|f\|$. – Pedro Nov 26 '18 at 15:51

However, I would recommend to specify the norms for $$f$$ and $$v$$ in your proof (I assume you mean the $$L^2(\Omega)$$ norm). For $$l$$ it is clear in my opinion that the norm refers to the operator norm.
Further, we can obtain $$\|l\|=\|f\|_{L^2(\Omega)}$$. This can be seen by choosing $$v:=f\in L^2(\Omega)$$. Then we have $$\| l \|\|f\|_{L^2(\Omega)} \geq |l(f)| = (f,f)_\Omega = \|f\|^2_{L^2(\Omega)}.$$ Dividing by $$\|f\|_{L^2(\Omega)}$$ yields the result (you have to consider the special case $$f=0$$, which is trivial).
• Thank you very much! Is there any theorem related to this result (i.e., $\|l\|=\|f\|$) ? I am thinking about the Riesz representation theorem that says, we can always find a unique $f \in H$ such that $l_f(v) = (f,v)$ for $\forall v \in H$ and $l \in H^*$, and $\|l\|=\|f\|$. So can we say our result is a consequence of the Riesz representation theorem? – Analysis Newbie Nov 26 '18 at 17:53