Matematické Fórum

Olin · 06. 07. 2011 11:25

Let $a, b$ be reals satisfying $1 < a < b$ and $f, g \colon [0, 1] \to [0, \infty)$ continuous functions satisfying

$\int_0^1 f^b(x) \, \mathrm{d}x \leq \int_0^1 f^a(x) g(x) \, \mathrm{d}x$ .

Prove that

$\int_0^1 f^b(x) \, \mathrm{d}x \leq \int_0^1 g^{\frac{b}{b-a}}(x) \, \mathrm{d}x$ .

laszky · 09. 03. 2018 02:24 — Editoval laszky (09. 03. 2018 03:56)

We use Holder's inequality $\int_0^1 |u(x)v(x)| \mathrm{d}x \leq \left(\int_0^1|u(x)|^p \mathrm{d} x\right)^{\frac{1}{p}}\left(\int_0^1|v(x)|^q \mathrm{d} x\right)^{\frac{1}{q}}$ which holds for all $p,q\geq1$ satisfying $\frac{1}{p}+\frac{1}{q}=1$ and $u\in L^p(0,1), v\in L^q(0,1)$ .

Hence, denoting $|u(x)|=f^a(x)$ , $|v(x)|=g(x)$ , $p=\frac{b}{a}$ and $q=\frac{b}{b-a}$ we may write

$\int_0^1 f^b(x) \, \mathrm{d} x \leq \int_0^1f^a(x)g(x) \,\mathrm{d} x \leq \left(\int_0^1f^b(x)\, \mathrm{d} x\right)^{\frac{a}{b}} \left(\int_0^1g^{\frac{b}{b-a}}(x)\, \mathrm{d} x\right)^{\frac{b-a}{b}}$ .

Dividing both sides of this inequality by $\left(\int_0^1f^b(x)\, \mathrm{d} x\right)^{\frac{a}{b}}$ yields

$\left(\int_0^1f^b(x)\, \mathrm{d} x\right)^{\frac{b-a}{b}}\leq \left(\int_0^1g^{\frac{b}{b-a}}(x)\, \mathrm{d} x\right)^{\frac{b-a}{b}}$ ,

which is the desired inequality.

Matematické Fórum

#1 06. 07. 2011 11:25

Easy inequality with integrals

#2 09. 03. 2018 02:24 — Editoval laszky (09. 03. 2018 03:56)

Re: Easy inequality with integrals

Zápatí