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 $\|l\|=\|f\|$?
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).