\( \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\P}{\mathcal P} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathbb{F}} \newcommand{\E}{\mathcal E} \newcommand{\brac}[1]{\left(#1\right)} \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\matrixx}[1]{\begin{bmatrix}#1\end {bmatrix}} \newcommand{\vmatrixx}[1]{\begin{vmatrix} #1\end{vmatrix}} \newcommand{\lims}{\mathop{\overline{\lim}}} \newcommand{\limi}{\mathop{\underline{\lim}}} \newcommand{\limn}{\lim_{n\to\infty}} \newcommand{\limsn}{\lims_{n\to\infty}} \newcommand{\limin}{\limi_{n\to\infty}} \newcommand{\nul}{\mathop{\mathrm{Nul}}} \newcommand{\col}{\mathop{\mathrm{Col}}} \newcommand{\rank}{\mathop{\mathrm{Rank}}} \newcommand{\dis}{\displaystyle} \newcommand{\spann}{\mathop{\mathrm{span}}} \newcommand{\range}{\mathop{\mathrm{range}}} \newcommand{\inner}[1]{\langle #1 \rangle} \newcommand{\innerr}[1]{\left\langle #1 \right \rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\toto}{\rightrightarrows} \newcommand{\upto}{\nearrow} \newcommand{\downto}{\searrow} \newcommand{\qed}{\quad \blacksquare} \newcommand{\tr}{\mathop{\mathrm{tr}}} \newcommand{\bm}{\boldsymbol} \newcommand{\cupp}{\bigcup} \newcommand{\capp}{\bigcap} \newcommand{\sqcupp}{\bigsqcup} \newcommand{\re}{\mathop{\mathrm{Re}}} \newcommand{\im}{\mathop{\mathrm{Im}}} \newcommand{\comma}{\text{,}} \newcommand{\foot}{\text{。}} \)

Friday, August 14, 2015

On Convex Functions

Today I came across a result for AC functions.

Theorem. If $f :I\to \R$ is absolutely continuous with $f'\in BV(I)$, then $f$ is representable as a difference of two convex functions.

We need this result for generalised Ito's Lemma which works on convex functions (with an indirect route: consideration of local time process). This result seems nonstandard, and I have paid a few hours finding standard results on convex functions, the following turns out to be what I want:

Lemma (Thm 14.14 of J. Yeh's Real analysis). Let $f$ be a real-valued function an open interval $I$ in $\R$. Suppose that
1) $f$ is AC on any closed subinterval of $I$; and
2) $f'$ is increasing on the subset, $A$, of $I$ on which $f'$ exists and $m(A)=m(I)$.
Then $f$ is a convex function on $I$.

Having the lemma we can prove the Theorem immediately.

Proof. Since $f$ is $AC$, fix $a\in I$ and for any $x\in I$ we have \[f(x)=f(a)+\int_a^xf'(s)\,ds.\] Now we can decompose $f'$ into a difference of two nondecreasing functions since $f'\in BV$, call them $H, K$, i.e., $f'=H-K$. As a result, \[
f(x) = f(a) +\int_a^x H\,ds - \int_a^x K\,ds,
\] finally we denote $h = \int_a^x H\,ds$ and $k = \int_a^x K\,ds$. Then $h,k$ are AC on any closed subinterval of $I$, moreover, $h' = H,k'=K$ a.e. and they are increasing, therefore we can apply the lemma to conclude that $h$ and $k$ are convex.$\qed$

There are many interesting and fundamental facts for convex functions that are not mentioned in UG curriculum of UST and I really suggesting reading them all, they are too standard to miss.

Still I want to record an important fact of convex functions: $f'_-(x)$ is increasing (of course) and always left continuous if $f$ is convex, this makes if possible to define Stieljes measure by using $f'_-$.

No comments:

Post a Comment