Random math: 32

Source of most content: Gunhee Cho and RCA by Rudin.


Def. For any open set $(a,b) \in [-\infty, \infty]$, $\varphi: (a,b) \to \mathbb{R}$ is convex if $\forall x,y \in (a,b)$ with $x < y$ and $\forall \lambda \in [0, 1]$, $\varphi(\lambda x + (1-\lambda) y) \leq \lambda \varphi(x) + (1- \lambda) \varphi(y)$ holds.

NB1; If we put $t=\lambda x + (1-\lambda) y$, it holds that $\varphi(t) \leq \lambda \varphi(x) + (1-\lambda)\varphi(y)$. Since $\lambda (x-y) = t-y$ holds, we have $\frac{\varphi(x)-\varphi(y)}{x-y} \leq \frac{\varphi(t)-\varphi(y)}{t-y}$.
NB2; More generally, for $a < s < t < u < b$, it holds that $\frac{\varphi(t)-\varphi(s)}{t-s} \leq \frac{\varphi(u)-\varphi(t)}{u-t}$.
NB3; Thus, a real differentiable function $\varphi$ is convex in $(a,b)$ iff $a< x < y < b$ implies $\varphi’(x) \leq \varphi’(y)$, i.e., the derivative $\varphi’$ is a monotonically increasing function.
NB4; If $\varphi$ is convex on $(a,b)$, then $\varphi$ is continous on $(a,b)$.
NB5; If $f, g$ are convex on $(a,b)$, then $\max (f, g)$ is convex on $(a,b)$.

Jensen’s Inequality

\[\text{Let }\mu \text{ be a measure on } (X, F) \text{ s.t. } \mu(X) =1. \text{ If a real fn } f \in L^1(\mu) \text{ s.t. } \forall x \in X, \, f(x) \in (a,b), \text{ and if } \varphi \text{ is cvx, } \nonumber\] \[\text{then } \varphi \left( \int_X f d\mu \right) \leq \int_X \varphi \circ f d\mu. \nonumber\]


Put $t= \int_X f d\mu \in (a,b)$. Let $\beta := \sup \{ \frac{\varphi(s)-\varphi(t)}{s-t} : a < s < t \} \in \mathbb{R}$. By LUP, it holds that $\forall s \in (a,b), $ $ \varphi(s) - \varphi(t) \geq \beta (s-t)$. Since $\forall x \in X, f(x) \in (a,b)$ holds by assumption, $\varphi(f(x)) - \varphi(t) \geq \beta (f(x) -t )$ holds. Since $\varphi$ is contnous by NB4, if we integrate w.r.t. $\mu$, we have $\int_X \varphi \circ f d\mu - \int_X \varphi(t) d \mu \geq \beta \int_X (f(x) -t) d\mu = \beta \int_X f d\mu - \beta\int_X t d \mu \stackrel{t=\int_X f d\mu}{=} 0$. Hence, we have $\int_X \varphi \circ f d\mu \geq \int_X \varphi(t) d \mu = \varphi(t) \int_X 1 d\mu \stackrel{\mu(X) =1}{=} \varphi(t) = \varphi \left( \int_X f d\mu \right)$, which concludes the proof.

NB6; Let $\varphi(x) = e^x$ (cvx). By Jensen’s Ineq, $\exp(\int_X f) \leq \int_X \exp f$ holds. Here, if we set $f(x) = \log g(x)$, we have $\exp (\int_X \log g) \leq \int_X g$.
NB7; Let $\mu (\{ \frac{1}{n}\}) = \frac{1}{n} $ and $f(x_i) = p_i\in \mathbb{R}$ for finite set $X= \{ x_1, \ldots, x_n \}$. Then, $\int_X f = \frac{1}{n}(p_1 + \cdots + p_n)$ holds. By applying Jensen’s Ineq, we have $\exp ( \frac{1}{n}(p_1 + \cdots + p_n) ) \leq \frac{1}{n}\left[ e^{p_1} + \cdots + e^{p_n} \right] $. Letting $e^{p_i} = y_i$ gives the Arithmetic-geometric inequality.
NB8; If $\mu(x_i) = \alpha_i $ s.t. $\sum \alpha_i = 1$, we have $ \Pi_i y_i^{\alpha_i} \leq \sum \alpha_i y_i$.

Hölder and Minkowski’s Inequalities

Def. Let $p , q \in \mathbb{R}_{+}$ s.t. $p+q = pq$, i.e., $\frac{1}{p} + \frac{1}{q} =1$. We say $(p,q)$ conjugate exponents.

NB9; $(1, \infty)$ and $(\infty , 1)$ are also considered as conjugate exponents.

\[\text{Given M.S. } (X, F, \mu) \text{, conjugate exponents } (p,q) \text{, measurable functions } f,g \in [0, \infty]. \text{ Then, }\nonumber\] \[\text{ Hölder's Ineq: } \int_X fg d\mu \leq \left( \int_X f^p d\mu \right)^{1/p} \left( \int_X g^q d\mu \right)^{1/q} \nonumber\] \[\text{Minkowski's Ineq: } \left(\int_X (f+g)^p d\mu \right)^{1/p} \leq \left( \int_X f^p d\mu \right)^{1/p} + \left( \int_X g^p d\mu \right)^{1/p} \nonumber\]


  • (Hölder) : Let $\left( \int_X f^p d\mu \right)^{1/p} = A$ and $\left( \int_X g^q d\mu \right)^{1/q} =B$. If $A=\infty$, it holds trivially and if $A=0$, then $\int_X f^p d\mu =0$ holds. Since $f\geq 0$, this implies that $f^p = 0$ almost everywhere. Hence, $fg=0$ holds almost everywhere, which gives $\int_X fg =0$. Therefore, assume $0 < A, B < \infty$ holds and set $F(x) := \frac{f(x)}{A}$ and $G(x) := \frac{g(x)}{B}$ so that $\int F^p = \frac{A^p}{A^p} =1$ and $\int G^p = \frac{B^p}{B^p} =1$. Since $F^p, G^p \ne 0, \infty$ a.e., we may assume that $0 < F(x), G(x) < \infty$ holds for all $x\in X$. Hence, for each $x \in X$, $\exists! s,t \in \mathbb{R} $ s.t. $F(x)=\exp \frac{s}{p}, \, G(x)=\exp \frac{t}{q}$ for $\frac{1}{p} + \frac{1}{q} = 1$. By convexity of $\exp$, we have $\exp \frac{s}{p} \exp \frac{t}{q} \leq \frac{1}{p}e^s + \frac{1}{q} e^t$, i.e., $FG \leq \frac{1}{p}F^p + \frac{1}{q}G^q$. Thus, $\int_X FG \leq \frac{1}{p} \int F^p + \frac{1}{q} \int G^q = \frac{1}{p} + \frac{1}{q} = 1$, which concludes the proof since this gives $\frac{1}{AB} \int_X fg \leq 1$.
  • (Minkowski) : Since $(p,q)$ is a conjugate exponent, $(p-1)q = p$ and $(q-1)p =q$ holds. As $(f+g)^p = (f+g)(f+g)^{p-1} = f(f+g)^{p-1} + g(f+g)^{p-1}$ holds, we have $\int_X f(f+g)^{p-1} \stackrel{\text{Hölder}}{\leq} \left( \int f^p \right)^{1/p} \left( \int (f+g)^{(p-1)/q} \right)^{1/q} = \left( \int f^p \right)^{1/p} \left( \int (f+g)^{p} \right)^{1/q}$. Thefore, $\int (f+g)^p = \int f(f+g)^{p-1} + g(f+g)^{p-1} \leq \left( \int (f+g)^{p} \right)^{1/q} \left( \left( \int f^p \right)^{1/p} + \left( \int g^p \right)^{1/p}\right)$ holds, i.e., $ \left(\int (f+g)^p\right)^{1-1/q} \leq \left( \int f^p \right)^{1/p} + \left( \int g^p \right)^{1/p}$.

$L_p$ space

Def. Given a M.S. $(X, F, \mu)$ for $0 < p < \infty$, define $L_p$ space $L^p(\mu) := \{ f: X \to \mathbb{C} : \left(\int_X \mid f \mid^p d \mu \right)^{1/p} < \infty \}$ where $f$ is complex measurable function on $X$. We call $\Vert f \Vert_p := \left(\int_X \mid f \mid^p d \mu \right)^{1/p}$ the $L_p$ norm of $f$.

NB10; When $p=1$, recall that $L^1(\mu)$ space is a collection of Lebesgue integrable measurable functions.
NB11; When $X$ is countable set and $\mu$ is a counting measure, we denote by $\ell^p(\mu) := \{ (x_n) \in \mathbb{C} : \sum_{n=1}^\infty \mid x_n \mid^p < \infty \}$.
NB12; Fix $p \in [1, \infty]$. If $f \in L^p(\mu)$ and $\alpha \in \mathbb{C}$, then $\alpha f \in L^p(\mu)$ holds clearly and $\Vert \alpha f \Vert_{p} = \mid \alpha \mid \Vert f \Vert_{p}$.

Def. Given a M.S. $(X, F, \mu)$, suppose $g: X \to [0, \infty]$ be measurable function on $X$. Let $S$ be the set of all real $\alpha$ s.t. $\mu(g^{-1}((\alpha, \infty]))=0$. Define $ \Vert g \Vert_{\infty} := \inf S$ and call essential bound (supremum) of $g$. And define $L^\infty(\mu) := \{ f: X \to \mathbb{C} : \Vert f \Vert_{\infty} < \infty\}$.

NB13; If $S=\emptyset$, then let $\inf S = \infty$.
NB14; If $S \ne \emptyset$, $\beta = \inf S \in \mathbb{R}$, then $g^{-1}((\beta, \infty]) = \cup_{n=1}^\infty g^{-1}((\beta+\frac{1}{n}, \infty])$ which is a measure zero set. Thus, $\mu(g^{-1}((\beta, \infty])) =0$, i.e., $\beta \in S$.
NB15; $L^p(\mu)$ is a complete normed linear space (Banach space) for $1\leq p \leq \infty$.


\[\text{Suppose } 1 \leq p \leq \infty \text{ for conjugate exponents } (p,q) \text{ and if } f \in L^p(\mu) \text{ and } g \in L^q (\mu). \nonumber\] \[\text{ Then } fg \in L^1(\mu) \text{ and } \Vert fg \Vert_{1} \leq \Vert f \Vert_{p} \Vert g \Vert_{q}. \nonumber\]


For $p \in (1, \infty)$, it is simply Hölder’s Ineq. which was shown. By definition, $ \mid fg \mid \leq \Vert f \Vert_{\infty} \mid g \mid$ holds a.e. and thus integrating this concludes the proof when $p=\infty$. The same argument holds for $p=1$ and $q=\infty$ holds.

\[\text{Suppose } 1 \leq p \leq \infty, \, f \in L^p(\mu) \text{ and } g \in L^q (\mu), \text{ then } f+g \in L^p(\mu) \text{ and } \Vert f+g \Vert_{p} \leq \Vert f \Vert_{p} \Vert g \Vert_{p}. \nonumber\]


For $p \in (1, \infty)$, it follows from Minkowski’s Ineq. which was shown above. For $p=1$ or $p=\infty$, it holds trivially since $\mid f+g \mid \leq \mid f \mid + \mid g \mid$ holds.