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A convex real function is Lipschitz on every closed bounded subinterval of the interior of its domain, hence continuous throughout the interior
Statement
Let be convex and let . Then there is such that for all . Thus is Lipschitz on (Lipschitz map, -Hölder map for rational , and contraction) and is continuous at every point of (Continuity of at a point of and on : the - condition, its agreement with at a limit point, and continuity at an isolated point).
Facts & Assumptions
Given: A convex function and .
For a convex function and , the three secant slopes satisfy (For a convex function and , the three secant slopes satisfy ).
A function is Lipschitz with constant when for all points in its domain (Lipschitz map, -Hölder map for rational , and contraction).
Proof
Choose with ; then for , two applications of the three-slope inequality give .
With , step 1.1 yields for ; symmetry gives the same estimate for all , which is the Lipschitz condition.
Given , choose such an interval containing in its interior; the estimate in step 2.1 gives the -- condition at by taking when , and is immediate when .
Depends on
- For a convex function and $x<y<z$, the three secant slopes satisfy $s(x,y)\le s(x,z)\le s(y,z)$
- Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point
- Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction
Used by
- Assuming Choice, a Hamel coefficient map is midpoint convex but discontinuous and therefore not convex Counterexample
- A convex function on an open interval has finite left and right derivatives everywhere, with f'_-(u)≤ f'_+(u)≤ (f(v)-f(u))/(v-u)≤ f'_-(v)≤ f'_+(v) for u<v Theorem
- A convex function on an open interval is differentiable except at at most countably many points Theorem
Dependency tree · two levels
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Sources
- R. Gardner, Convex Functions, Notes 6.6 (standard reference, not scraped)