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 $\|l\|=\|f\|$?

  • $\begingroup$ $\|\ell\|\leq \|f\|$ 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 $E$. $\endgroup$ – Surb Nov 26 '18 at 14:58
  • $\begingroup$ 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, $\|l\|\leq \|f\|$. $\endgroup$ – Pedro Nov 26 '18 at 15:51

Yes, this is correct.

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).

  • $\begingroup$ 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? $\endgroup$ – Analysis Newbie Nov 26 '18 at 17:53
  • $\begingroup$ yes, I would say this is the case. $\endgroup$ – supinf Nov 27 '18 at 10:13

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