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Young不等式 - Wikipedia

Young不等式

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数学上,-{Young}-不等式,指出:假设 a, b, pq 是正实数 ,且有1/p + 1/q = 1 ,那么:

ab \le \frac{a^p}{p} + \frac{b^q}{q}.
等号成立当且仅当 ap = bq ,因为这时ab = a(b^q)^{1 \over q} = aa^{p \over q} = a^p = {a^p \over p} + {b^q \over q}

-{Young}-不等式是加权算术-几何平均值不等式的特列,-{Young}-不等式是证明Hölder不等式的一个快捷方法。

[编辑] 证明

我们知道函数f(x) = ex 是一个凸函数, 因为它的二阶导数恒为正。 从而我们有:

ab = e^{\ln(a)}e^{\ln(b)} = e^{{1 \over p}\ln(a^p) + {1 \over q}\ln(b^q)} \le {1 \over p}e^{\ln(a^p)}+{1 \over q}e^{\ln(b^q)} = {a^p \over p} + {b^q \over q}.

这里我们使用了凸函数的一个性质:对任意 t ,若 0 < t <1,则有:

f(tx+(1-t)y)\leq t f(x)+(1-t)f(y)

[编辑] 推广

\phi : \mathbb{R} \rightarrow \mathbb{R}是一个连续、严格递增函数且 φ(0) = 0 。那么下面的不等式成立:

ab \leq \int_{0}^a \phi(x) dx + \int_{0}^b \phi^{-1}(y) dy

观察φ(x)的图形,很容易看出这个不等式的一个直观证明:以上两个积分式所表示的区域之和比由ab组成的三角形的面积大。

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