Lecture 13 - Inequalities

#math

Definition

$ϕ : (a, b) \subset R \to R$

is convex if $ϕ (λ x + μ y) \leq λ ϕ (x) + μ ϕ (y)$ for any $x, y \in (a, b)$ and $λ, μ \in R$ such that $λ + μ = 1$ .
A convex function is necessarily continuous.

Proposition

If $ϕ \in C^{2} (I)$ , then $ϕ$ is convex iff $ϕ^{″} \geq 0$ .

Young's Inequality

If $a, b \geq 0$ , $1 < p, q < \infty$ and $\frac{1}{p} + \frac{1}{q} = 1$ or $p + q = p q$ ( $q$ is said to be the conjugate of $p$ ), then:

a b \leq \frac{a^{p}}{p} + \frac{b^{q}}{q} .

Proof

If $a$ or $b$ is 0 then there is nothing to prove. So let $a, b > 0$ . We note $e^{x}$ is a convex function, and:

\begin{matrix} ◻ & a b = e^{l n (a b)} = e^{l n (a) + l n (b)} = e^{\frac{1}{p} l n (a^{p}) + \frac{1}{q} l n (b^{q})} \leq \frac{1}{p} e^{l n (a^{p})} + \frac{1}{q} e^{l n (b^{q})} = \frac{1}{p} a^{p} + \frac{1}{q} b^{q} \end{matrix}

Corollary

If $a, b > 0$ , $1 < p, q < \infty$ such that $p + q = p q$ , and $ϵ > 0$ there holds:

a b \leq \frac{ϵ}{p} a^{p} + \frac{1}{q ϵ^{q / p}} b^{q}

Proof

$a b = ϵ^{1 / p} a \frac{b}{ϵ^{1 / p}}$

Holder's Inequality

If $u, v$ are measurable functions on $Ω \in R^{N}$ , $1 \leq p, q \leq \infty$ and $\frac{1}{p} + \frac{1}{q} = 1$ then:

\int_{Ω} | u v | \leq {(\int_{Ω} | u |^{p})}^{\frac{1}{p}} {(\int_{Ω} | v |^{q})}^{\frac{1}{q}}

Proof

We assume that $| | u | |_{p}$ and $| | v | |_{q}$ are finite and not zero. When $p = q = 1$ or $\infty$ the proof is simple. We assume $1 < p, q < \infty$ . We then apply the corollary above to $| u (x) |$ and $| v (x) |$ and integrate over $Ω$ :

\int_{Ω} | u (x) | | v (x) | d x \leq \frac{ϵ}{p} \int_{Ω} | u (x) |^{p} d x + \frac{1}{q ϵ^{q / p}} \int_{Ω} | v (x) |^{q} d x

By choosing $ϵ = {(\frac{\int_{Ω} | v |^{q}}{\int_{Ω} | u |^{p}})}^{\frac{1}{q}}$ the right hand side of the above inequality gives us Holder's inequaity.

When $p = q = 2$ , Holder's Inequality is equivalent to the Cauchy-Schwarz Inequality.

Minkowski Inequality

If $u, v$ are measurable functions on $Ω \subset R^{N}$ and $1 \leq p \leq \infty$ , then:

| | u + v | |_{L^{p} (Ω)} \leq | | u | |_{L^{p} (Ω)} + | | v | |_{L^{p} (Ω)}

Discrete version: let $\frac{1}{p} \frac{+ 1}{q} = 1$ , then:

\sum_{k} | a_{k} | | b_{k} | \leq {(\sum_{k} | a_{k} |^{p})}^{\frac{1}{p}} {(\sum_{k} | b_{k} |^{q})}^{\frac{1}{q}}

(\sum_{k} | a_{k} + b_{k} |^{p})^{\frac{1}{p}} \leq {(\sum | a_{k} |^{p})}^{\frac{1}{p}} + {(\sum | b_{k} |^{q})}^{\frac{1}{q}}

Green's Identity

$\int_{Ω} (u Δ v - w Δ u) = \int_{δ Ω} \frac{\partial w}{\partial η} - w \frac{\partial u}{\partial η}$

If $u \equiv 1$ then $Δ u = 0$ , giving us:

\int_{Ω} Δ w = \int_{\partial Ω} \frac{\partial w}{\partial η}

We want to solve:

\begin{matrix} (*) & - Δ w = f (x), {\frac{\partial w}{\partial η} |}_{\partial Ω} = 0 \end{matrix}

Note: ${\frac{\partial w}{\partial η} |}_{\partial Ω} = 0$ means the directional derivative restricted to the boundary is equivalently 0.
However integrating over $Ω$ we get:

0 = - \int_{\partial Ω} \frac{\partial w}{\partial η} = - \int_{Ω} Δ w = \int_{Ω} f (x)

So $(*)$ can only be solved if $\int_{Ω} f (x) = 0$ .