Hölder's Inequality for Integrals

Hölder Inequality (검색하기 힘들고, 무엇보다 ö를 타이핑치는 일이 상당히 귀찮으므로 Holder라고 쓰려고 한다), 

횔더 부등식 (이렇게 읽는게 맞나?) 은 가장 elementary 한 형태로는 다음과 같은 형태의 부등식이다.

$\frac{1}{p} + \frac{1}{q} = 1$ 일 때,

$$\sum_{k = 1} ^ {n} |x_k y_k| \leq \left( \sum_{k = 1}^{n} |x_k|^p \right)^{\frac{1}{p}} \left( \sum_{k = 1}^{n} |y_k|^q \right)^{\frac{1}{q}}$$

이 형태를 증명하는 것도 크게 다르지 않으니, 우리는 적분 형태로 기술된 Holder 부등식을 증명해 보자. 적분으로 기술된 Holder 부등식은 다음과 같다. ($p, q$ 에 대한 조건은 위와 같다.) 아래 모든 함수들은 리만-스틸체스 적분 가능하고 가정하고, $\alpha$ 는 적당히 단조증가인 함수를 골랐다고 하자.

$$\int_{a}^{b} |f(x)g(x)| d\alpha \leq \left(\int_{a}^{b} |f(x)|^p d\alpha \right)\left(\int_{a}^{b} |g(x)|^q d\alpha \right)$$

먼저, 위 조건 (역수의 합이 1) 을 만족하는 $p, q$ 에 대하여, 다음 Young's Inequality를 보이자.

[Young's Inequality]
임의의 양수 $x, y$에 대해, $\frac{1}{p} + \frac{1}{q} = 1$ 일 때, $$xy \leq \frac{x^p}{p} + \frac{y^q}{q}$$

[증명]
함수 $f(x) = - \log{x}$ 를 생각하자. 이 함수는 두 번 미분 가능하고, $f''(x)$ 는 항상 음수이므로 convex하다. 이제, convex 한 함수에서는 $f(t x + (1-t)y) \geq tf(x)+ (1-t)f(y)$ 임을 이용하면, 다음과 같이 쓸 수 있다.
$$-\log{\left(\frac{x^p}{p} + \frac{y^q}{q}\right)} \leq - \frac{1}{p} \log{x^p} - \frac{1}{q} \log{y^q}$$
로그의 기본 성질을 이용하면,
$$-\log{\left(\frac{x^p}{p} + \frac{y^q}{q}\right)} \leq -\log{x} - \log{y}$$ 이므로, $\log{xy} \leq \log{\left(\frac{x^p}{p} + \frac{y^q}{q}\right)}$ 가 되어 위 부등식이 성립한다.

이제, 다음과 같이 $u(x), v(x)$ 를 정의하자.
$$u(x) = \frac{|f(x)|}{\left(\int_a^b |f(x)|^p \ d\alpha\right)^\frac{1}{p}}$$

$$v(x) = \frac{|g(x)|}{\left(\int_a^b |g(x)|^q \ d\alpha\right)^\frac{1}{q}}$$

자명하게, 두 함수 모두 항상 양수이므로 모든 점에서 Young's Inequality 가 성립한다.
$$u(x)v(x) \leq \frac{1}{p} u(x)^p + \frac{1}{q} v(x)^q$$
$u$와 $v$를 풀어서 $f$와 $g$에 대해 쓰면 다음을 얻는다.

$$\frac{|f(x)||g(x)|}{\left(\int_a^b |f(x)|^p \ d\alpha\right)^\frac{1}{p}\left(\int_a^b |g(x)|^q \ d\alpha\right)^\frac{1}{q}} \leq \frac{|f(x)|^p}{\left(p \int_a^b |f(x)|^p \ d\alpha\right)}+\frac{|g(x)|^q}{\left(q \int_a^b |g(x)|^q \ d\alpha\right)}$$

양변을 $a$ 부터 $b$ 까지 $\alpha$에 대해서 정적분하자. 그러면, 우변에서 적분식이 모두 날아갈 것이므로 (이미 정적분된 값은 어떤 상수일 것이다.)
$$\frac{\int_a^b |f(x)||g(x)| d\alpha}{\left(\int_a^b |f(x)|^p \ d\alpha\right)^\frac{1}{p}\left(\int_a^b |g(x)|^q \ d\alpha\right)^\frac{1}{q}} \leq \frac{1}{p} + \frac{1}{q} = 1$$

우리가 원하는 식을 얻는다.
$$\int_{a}^{b} |f(x)g(x)| d\alpha \leq \left(\int_{a}^{b} |f(x)|^p d\alpha \right)\left(\int_{a}^{b} |g(x)|^q d\alpha \right)$$