+++ title = "Inequality" description = "Sometimes 「inequality」 is more important than 「equation」" categories = ["MATH","数学基础"] tags = [""] keywords = ["不等式","inequality","Cauchy inequality","Young's inequality","Cauchy inequality for scalar product","Holder inequality","FEM"] date = "2020-03-17T00:21:47+08:00" toc = true mathjax = true aliases = ["/posts/fem/inequality"] +++
Let
\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.
Constant deformation with
Let
\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 $$ \langle \mathbf{x}, \mathbf{y} \rangle ^2 \leq \lVert \mathbf{x} \rVert^2 \lVert \mathbf{y} \rVert^2 $$
For
\begin{equation} \label{eq:eqy} ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}. \quad (Young\ inequality) \end{equation}
Proof. The function
When
Let
\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}
{{< imgcap src="https://imgkr.cn-bj.ufileos.com/a41e796b-adbb-44a7-adbe-3da2bb474641.svg" title="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
While
Let
\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