Inequality

Sometimes 「inequality」 is more important than 「equation」

Cauchy inequality

Let $a, b$ are any real numbers, then

\begin{equation} \label{eq:eqc} ab \leq \frac{1}{2}(a^{2} + b^{2}),\quad (Cauchy\ inequality) \end{equation}

\begin{equation} \label{eq:eqce} ab \leq \epsilon a^{2} + \frac{b^{2}}{4\epsilon},\,\forall \epsilon > 0,\quad (Cauchy\ inequality\ with\ epsilon) \end{equation}

Proof.

$$ (a-b)^{2} \geq 0 \Rightarrow ab \leq\frac{1}{2}(a^{2} + b^{2}). $$

Constant deformation with $\epsilon$

$$ \begin{aligned} ab &= ((2\epsilon)^{\frac{1}{2}}a)\bigg(\frac{b}{(2\epsilon)^{\frac{1}{2}}}\bigg) \\[3pt] &\leq \frac{1}{2} \biggl[((2\epsilon)^{\frac{1}{2}}a)^2 + \bigg(\frac{b}{(2\epsilon)^{\frac{1}{2}}}\biggl)^2 \biggl] \\[3pt] &= \frac{1}{2} \biggl(2\epsilon a^{2} + \frac{b^{2}}{2\epsilon}\biggr) \\[3pt] &= \epsilon a^{2} + \frac{b^{2}}{4\epsilon}. \end{aligned} $$

$\blacksquare$

Cauchy inequality for scalar product

Let $\lVert \mathbf{x} \rVert^2 = \sum_{k=1}^n \lvert x \rvert^2$, $\langle \mathbf{x}, \mathbf{y} \rangle = \sum_{k=1}^n x_k \bar{y}_k$, which $\mathbf{x} = \{ x_k \}_{k=1,\ldots,n}, \mathbf{y} = \{ y_k \}_{k=1,\ldots,n}$ then

\begin{equation} \langle \mathbf{x}, \mathbf{y} \rangle \leq \lVert \mathbf{x} \rVert \lVert \mathbf{y} \rVert \end{equation}

Proof. We have $$ f(\lambda) = \lVert \mathbf{x} + \lambda\mathbf{y}\rVert^2 \geq 0, \quad \forall \lambda \in \mathbb{R} $$ Thus the quadratic function $$ f(\lambda) = \lVert \mathbf{y} \rVert^2 \lambda^2 + 2\langle \mathbf{x}, \mathbf{y} \rangle \lambda + \lVert \mathbf{x} \rVert^2 \geq 0 $$ According to the discriminant of the quadratic polynomial $$ \langle \mathbf{x}, \mathbf{y} \rangle ^2 \leq \lVert \mathbf{x} \rVert^2 \lVert \mathbf{y} \rVert^2 $$

$\blacksquare$

Young's inequality

For $a, b \geq 0$ and $p,q \geq 1$ such that $ \frac{1}{p} + \frac{1}{q} = 1$, then

\begin{equation} \label{eq:eqy} ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}. \quad (Young\ inequality) \end{equation}

Proof. The function $x \mapsto e^x$ is convex. This means that for $x, y\in \mathbb{R}$ and $0\leq \theta \leq 1$

$$ \mbox{exp}(\theta x + (1-\theta)y) \leq \theta\mbox{exp}(x) + (1-\theta)\mbox{exp}(y) $$

When $a$ or $b$ are zero the inequality is trivial. Otherwise, the theorem is just this inequality with

$$ x:= p \ln a \quad y:= q \ln b \quad \theta = \frac{1}{p} $$

$\blacksquare$

General Young’s inequality

Let $c>0$ and $f:[0, c] \to \mathbb{R}$ be a strictly increasing continuous function such that $f(0)=0$. Let $a\in [0, c]$ and $b\in [0, f(c)]$. Then

\begin{equation} ab \leq \int_0^a f(x) \, \mathrm{d}x + \int_0^b f^{-1}(x) \, \mathrm{d}x \label{eq:eqy2} \end{equation}

Figure 1 ◎ Figure 1

(\ref{eq:eqy2}) can be understood with the figure 1.

Lemma Let $c>0$. For a strictly increasing continuous function $f:[0, c]\to \mathbb{R}$ and $b\in [0, f(c)]$, we have that $$ bf^{-1}(b) = \int_0^b f^{-1}(x) \, \mathrm{d}x + \int_0^{f^{-1}(b)} f(x) \, \mathrm{d}x $$ Proof. It is enough to prove it when $f$ is $C^1$, as then a continuous $f$ can be uniformly approximated by strictly increasing $C^1$ functions while $f ^{−1}$ is also uniformly approximated. It is also enough to prove it when $f(a) \geq b$, as otherwise we can just interchange both $f, f^{−1}$ and $a, b$.

To prove it in this case we can change variables in the integral $\int_0^b\ f^{−1}(x)\, \mathrm{d}x$ by putting $x = f(y)$, and then integrate by parts: $$ \int_0^b f^{-1}(x) \, \mathrm{d}x = \int_0^{f^{-1}(b)} yf'(y) \, \mathrm{d}y = bf^{-1}(b) - \int_0^{f^{-1}(b)} f(x) \, \mathrm{d}x $$

$\blacksquare$

Proof. Use (\ref{eq:eqy2}), and $f(a)\geq b$, we get

$$ \begin{aligned} ab & = bf^{-1}(b) + b(a-f^{-1}(b)) \\[3pt] & = \int_0^b f^{-1}(x) \, \mathrm{d}x + \int_0^{f^{-1}(b)} f(x) \, \mathrm{d}x + b(a-f^{-1}(b)) \\[3pt] & \leq \int_0^b f^{-1}(x) \, \mathrm{d}x + \int_0^{f^{-1}(b)} f(x) \, \mathrm{d}x + \int_{f^{-1}(b)}^a f(x) \, \mathrm{d}x \\[3pt] & = \int_0^b f^{-1}(x) \, \mathrm{d}x + \int_0^a f(x) \, \mathrm{d}x \end{aligned} $$

While $f(a)\leq b$, the proof process is similar.

$\blacksquare$

Holder inequality

Let $p, q$ are real numbers $1\leq p, q\leq \infty$ connected by the relationship $\frac{1}{p} + \frac{1}{q} = 1$ and $u \in L^p, v \in L^q$. then

\begin{equation} \int \lvert uv \rvert \,\mathrm{d}x \leq \sqrt[p]{\int \lvert u \rvert \,\mathrm{d}x} \sqrt[q]{\int \lvert v \rvert \,\mathrm{d}x} \label{eq:eqh} \end{equation}

Proof. The formula Young inequality (\ref{eq:eqy}) implies $$ \int \lvert ab \rvert \,\mathrm{d}x \leq \frac{1}{p} \int \lvert a \rvert^p \,\mathrm{d}x + \frac{1}{q} \int \lvert b \rvert^q \,\mathrm{d}x $$

set $a=\frac{|u|}{c_1}, a=\frac{|v|}{c_2}$, where $c_1=\sqrt[p]{\int \lvert u \rvert \,\mathrm{d}x}$ and $\sqrt[q]{\int \lvert v \rvert \,\mathrm{d}x}$, then

$$ \begin{aligned} \frac{1}{c_1c_2} \int \lvert uv \rvert \,\mathrm{d}x & \leq \frac{1}{p} \int \Biggl| \frac{|u|}{c_1} \Biggr| ^p \,\mathrm{d}x + \frac{1}{q} \int \Biggl| \frac{|v|}{c_2} \Biggr| ^q \,\mathrm{d}x \\[3pt] & = \frac{1}{p} + \frac{1}{q} \\[3pt] & = 1 \end{aligned} $$ Hence $$ \int \lvert uv \rvert \,\mathrm{d}x \leq c_1 c_2 = \sqrt[p]{\int \lvert u \rvert \,\mathrm{d}x} \sqrt[q]{\int \lvert v \rvert \,\mathrm{d}x} $$

$\blacksquare$

updatedupdated2020-04-042020-04-04
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