----- > [!proposition] Proposition. ([[convex image of Ax + b is a convex function]]) > Let $A$ be an $m$-by-$n$ [[matrix]] and $b \in \rr ^{m}$. For a [[convex function]] $f: \rr ^{m} \to \rr$, the function $g(x)=f(Ax+b)$ is [[convex function|convex]]. > [!proposition] Corollary. > Since [[norms are convex functions]], $\|Ax+b\|$ a convex function of $x$ for any norm $\|\cdot\|$. > [!proof]- Proof. ([[convex image of Ax + b is a convex function]]) > Define $h: \rr ^{n} \to \rr ^{m} \text{ by } h(x)=Ax+b$, so that $g(x)=f\big(h(x)\big)$. Then for arbitrary $t \in [0,1]$ and $u,v \in \rr ^{n}$ we have $\begin{align}g(tu+(1-t)v) = & f\big(h(tu+(1-t)v)\big) \\ = & f\big(A(tu+(1-t)v) + b + 0\big) \\ = & f\big(A(tu+(1-t)v) + b + bt-bt\big) \\ = & f\big(Atu + Av - Atv + b + bt - bt\big) \\ = & f\big(\textcolor{Thistle}{tAu +bt} + \textcolor{Apricot}{Av+b-tAv-bt} \big) \\ = & f\big(t(\textcolor{Thistle}{Au+b}) + (1-t)\textcolor{Apricot}{(Av+b)}\big) \\ \leq & tf\big(\textcolor{Thistle}{Au+b}\big) + (1-t) f\big(\textcolor{Apricot}{Av+b}\big) \\ = & tg(\textcolor{Thistle}{u}) + (1-t)g(\textcolor{Apricot}{v}). \end{align}$ By definition this means $g$ is [[convex function|convex]]. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```