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Chapter 3 Geometry of convex functions - Meboo Publishing ...

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3.6. EPIGRAPH, SUBLEVEL SET 233<br />

q(x)<br />

f(x)<br />

quasi<strong>convex</strong><br />

<strong>convex</strong><br />

x<br />

x<br />

Figure 73: Quasi<strong>convex</strong> function q epigraph is not necessarily <strong>convex</strong>, but<br />

<strong>convex</strong> function f epigraph is <strong>convex</strong> in any dimension. Sublevel sets are<br />

necessarily <strong>convex</strong> for either function, but sufficient only for quasi<strong>convex</strong>ity.<br />

3.6 Epigraph, Sublevel set<br />

It is well established that a continuous real function is <strong>convex</strong> if and<br />

only if its epigraph makes a <strong>convex</strong> set. [195] [301] [351] [364] [244]<br />

Thereby, piecewise-continuous <strong>convex</strong> <strong>functions</strong> are admitted. Epigraph is<br />

the connection between <strong>convex</strong> sets and <strong>convex</strong> <strong>functions</strong>. Its generalization<br />

to a vector-valued function f(X) : R p×k →R M is straightforward: [289]<br />

epi f {(X , t)∈ R p×k × R M | X ∈ domf , f(X) ≼<br />

t } (534)<br />

R M +<br />

id est,<br />

f <strong>convex</strong> ⇔ epif <strong>convex</strong> (535)<br />

Necessity is proven: [59, exer.3.60] Given any (X, u), (Y , v)∈ epif , we<br />

must show for all µ∈[0, 1] that µ(X, u) + (1−µ)(Y , v)∈ epif ; id est, we<br />

must show<br />

f(µX + (1−µ)Y ) ≼<br />

µu + (1−µ)v (536)<br />

R M +<br />

Yet this holds by definition because f(µX+(1−µ)Y ) ≼ µf(X)+(1−µ)f(Y ).<br />

The converse also holds.

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