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[parent] convex functions lie above their supporting lines (Result)

Let $f:\mathbf{R}\rightarrow \mathbf{R}$ be a convex, twice differentiable function on $[a,b]$ . Then $f(x)$ lies above its supporting lines, i.e. it's greater than any tangent line in $[a,b]$ .

Proof. :

Let $r(x)=f\left( x_{0}\right) +f^{\prime }\left( x_{0}\right) \left( x-x_{0}\right) $ be the tangent of $f(x)$ in $x=x_{0}\in \lbrack a,b].$

By Taylor theorem, with remainder in Lagrange form, one has, for any $x\in \lbrack a,b]$ : $$ f\left( x\right) =f\left( x_{0}\right) +f^{\prime }\left( x_{0}\right) \left( x-x_{0}\right) +\frac{1}{2}f^{^{\prime \prime }}\left( \xi \left( x\right) \right) \left( x-x_{0}\right) ^{2} $$ with $\xi \left( x\right) \in \lbrack a,b]$ . Then $$ f\left( x\right) -r(x)=\frac{1}{2}f^{^{\prime \prime }}\left( \xi \left( x\right) \right) \left( x-x_{0}\right) ^{2}\geq 0 $$ since $f^{^{\prime \prime }}\left( \xi \left( x\right) \right) \geq 0$ by convexity. $ \qedsymbol$




"convex functions lie above their supporting lines" is owned by Andrea Ambrosio.
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Cross-references: remainder, Taylor theorem, tangent line, lines, function, twice differentiable, convex
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This is version 2 of convex functions lie above their supporting lines, born on 2007-04-26, modified 2007-05-23.
Object id is 9268, canonical name is ConvexFunctionsLieAboveTheirSupportingLines.
Accessed 1032 times total.

Classification:
AMS MSC26B25 (Real functions :: Functions of several variables :: Convexity, generalizations)
 26A51 (Real functions :: Functions of one variable :: Convexity, generalizations)
 52A41 (Convex and discrete geometry :: General convexity :: Convex functions and convex programs)

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