The basis-pursuit denoising objective and residual indicator are proper closed convex

The function $f$ is finite everywhere because it is a finite sum of absolute values, and $f(0)=0$, so its effective domain is nonempty. For $0\le \theta \le 1$, the triangle inequality and homogeneity in [F1] give $f((1-\theta )x+\theta z)\le f((1-\theta )x)+f(\theta z)=(1-\theta )f(x)+\theta f(z)\mathrm{.}$ Thus $f$ is proper and convex by [F2,F3].

For $(x,t)\in {\mathbb{R}}^{n+1}$ outside $\mathrm{epi}f$, the gap $d=f(x)-t$ is positive. Put $L=\sqrt{n}+1>0$. If $(x^{\mathrm{\prime }},t^{\mathrm{\prime }})$ has product Euclidean distance $D<d\mathrm{/}(2L)$ from $(x,t)$, then coordinate-square domination gives $\mathrm{\parallel }{x}^{\mathrm{\prime }}-x{\mathrm{\parallel }}_2\le D$ and $\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }\le D$. By [F1], $f(x^{\mathrm{\prime }})-t^{\mathrm{\prime }}\ge f(x)-t-\mathrm{\mid }f(x^{\mathrm{\prime }})-f(x)\mathrm{\mid }-\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }\ge d-(\sqrt{n}+1)D>d\mathrm{/}2>0.$ So the complement of $\mathrm{epi}f$ is open and its epigraph is closed. The scalar $L$ is positive by [F8]; the coordinate-square domination follows by comparing a partial sum with the full nonnegative sum and using [F9].

The residual ball $C$ contains $y$ because $ϵ\ge 0$. If $u,v\in C$ and $0\le \theta \le 1$, homogeneity and the triangle inequality in [F1] give $\mathrm{\parallel }(1-\theta )u+\theta v-y{\mathrm{\parallel }}_2\le (1-\theta )\mathrm{\parallel }u-y{\mathrm{\parallel }}_2+\theta \mathrm{\parallel }v-y{\mathrm{\parallel }}_2\le ϵ,$ so $C$ is convex. If $u\mathrm{\notin }C$, the gap $d=\mathrm{\parallel }u-y{\mathrm{\parallel }}_2-ϵ>0$ and the reverse triangle inequality show that every $v$ with $\mathrm{\parallel }v-u{\mathrm{\parallel }}_2<d\mathrm{/}2$ also lies outside $C$. Thus $C$ is closed.

By [F5], $\mathrm{epi}g=C\times [0,\mathrm{\infty })$. If $(u,t)$ is outside this epigraph, either $u\mathrm{\notin }C$ or $t<0$. In the first case step 1.3 supplies an open ball about $u$ disjoint from $C$, hence a product neighborhood disjoint from the epigraph; in the second case an interval about $t$ contained in $(-\mathrm{\infty },0)$ supplies such a product neighborhood. Therefore $\mathrm{epi}g$ is closed. Since $C$ is nonempty by step 1.3, $g$ is proper. If both endpoints lie in $C$, their convex combination lies in $C$ by step 1.3 and $g$ is zero there; if either endpoint has value $+\mathrm{\infty }$ with positive coefficient, the convexity inequality has right side $+\mathrm{\infty }$. Endpoint coefficients reduce to equality at the endpoint with coefficient one. Thus $g$ is convex, and [F2] proves that both functions have the asserted proper, closed, convex properties.