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Suppose that te function $f:[0,\infty)\rightarrow\mathbb{R}$ is continuous and strictly increasing, with $f(0)=0$ and $f([0,\infty))=[0,\infty)$. Then define $$F(x)=\int_{0}^{x}f\qquad\text{and}\qquad G(x)=\int_{0}^{x}f^{-1}\qquad\text{for all $x\geq 0$}$$ Prove Young's Inequality: $$ab\leq F(a)+G(b)\qquad\text{for all $a\geq0$ and $b\geq0$}$$

For this question, I think I need to use this formula:

Prove $\int_{0}^{x}f+\int_{0}^{f(x)}f^{-1}=xf(x)\qquad\text{for all $x\geq0$}$

I tried to use Cauchy Inequality, but it didn't give me any clews keep going to solve the problem. Can someone give me a hint or suggestion to start? Thanks.

Community
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Simple
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  • Look at the picture in the answer here: http://math.stackexchange.com/questions/149901/geometric-interpretation-of-youngs-inequality –  Feb 03 '16 at 14:04

2 Answers2

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Consider the function $\phi(b)=ab-\int_0^af-\int_0^bf^{-1}$. Then: $$\phi'(b)=a-f^{-1}(b)$$ Therefore $\phi(b)$ has an extremum at $b=f(a)$ where $\phi(b)=0$ using the formula you mentioned in the question. Also $\phi''(b) = -f^{-1'}(b)<0$ since $g'(f(x))f'(x)=1$ and $f'(x)>0$ (strictly increasing) where $g=f^{-1}$. Therefore, this function $\phi(b)$ has a maximum value of $0 \implies \phi(b)\leq0$ which is simply the inequality to be proved.

zed111
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We can use the fact that \begin{align*} \int_0^x f(u) \, du + \int_0^{f(x)} f^{-1}(u)\,du = x f(x), \end{align*} that is, \begin{align*} \int_0^x f(u)\, du + \int_0^x u \, df(u) = x f(x), \end{align*} which can be derived without the differentiability assumption.

Case 1, $f^{-1}(b) \ge a$. Then \begin{align*} \int_0^a f(u)\, du + \int_0^b f^{-1}(u)\, du &=\int_0^a f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u)\\ &=\int_0^a f(u)\, du+\int_0^a u\, df(u) + \int_a^{f^{-1}(b)} u\, df(u)\\ &\ge \int_0^a f(u)\, du+\int_0^a u\, df(u) + \int_a^{f^{-1}(b)} a\, df(u)\\ &= af(a) + a \big[b -f(a) \big]\\ &= ab. \end{align*}

Case 2, $f^{-1}(b) \le a$. Then \begin{align*} \int_0^a f(u)\, du + \int_0^b f^{-1}(u)\, du &=\int_0^a f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u)\\ &=\int_0^{f^{-1}(b)} f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u) + \int_{f^{-1}(b)}^a f(u)\, du\\ &\ge \int_0^{f^{-1}(b)} f(u)\, du+\int_0^{f^{-1}(b)} u\, df(u) + \int_{f^{-1}(b)}^a b\, du\\ &= f^{-1}(b) b + b \Big[a -f^{-1}(b) \Big]\\ &= ab. \end{align*}

Gordon
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