1. Young's Inequality

Hello,

Could someone please give me help with the following problem?

Suppose f is a continuous increasing function with f(0) = 0. Prove that for a,b > 0, we have a form of Young's Inequality,
$ab \leq \int_0^af(x)dx + \int_0^b f^{-1}(x)dx$
and that equality holds if and only if b = f(a)

Thanks!

2. Write the second integral as $\int_0^b\!\!f^{-1}(y)\,dy$, and make the substitution $y = f(x)$. Then it becomes $\int_0^{f^{-1}(b)}\!\!xf'(x)\,dx$. Integrate that by parts, to get $\int_0^b\!\!f^{-1}(y)\,dy = \Bigl[xf(x)\Bigr]_0^{f^{-1}(b)} - \int_0^{f^{-1}(b)}\!\!f(x)\,dx = b{f^{-1}(b)} - \int_0^{f^{-1}(b)}\!\!f(x)\,dx$.

Now suppose that b = f(a). Then the previous equation says that $ab = \int_0^a\!\!f(x)\,dx + \int_0^b\!\!f^{-1}(x)\,dx$. Thus equality holds for the desired inequality in that case.

Next, suppose that b > f(a). By the result in the "equality" case, we know that $af(a) = \int_0^a\!\!f(x)\,dx + \int_0^{f(a)}\!\!f^{-1}(x)\,dx$. But $f^{-1}$ is an increasing function, taking its minimum value a at the left end of the interval [f(a),b]. So $\int_{f(a)}^b\!\!f^{-1}(x)\,dx > a(b-f(a))$. Hence \begin{aligned}ab = af(a) + a(b-f(a)) &< \int_0^a\!\!f(x)\,dx + \int_0^{f(a)}\!\!f^{-1}(x)\,dx + \int_{f(a)}^b\!\!f^{-1}(x)\,dx \\ &= \int_0^a\!\!f(x)\,dx + \int_0^b\!\!f^{-1}(x)\,dx.\end{aligned}

A similar argument shows that the strict inequality also holds when b < f(a).

[But I don't see how this has anything to do with Young's inequality (except that they both say that ab is less than something).]

3. Where do $f(x)$ is increasing and $f(0)=0$ come into play?

Is it to show $f(x)$ has a continuous inverse?

4. Originally Posted by chiph588@
Where do $f(x)$ is increasing and $f(0)=0$ come into play?

Is it to show $f(x)$ has a continuous inverse?
The condition that f is increasing is used in two ways. First, to ensure that f has a continuous, increasing inverse; and second, in the last part of the proof, to get the estimates for the integrals $\int_{f(a)}^b\!\!f^{-1}(x)\,dx > a(b-f(a))$ and (in the case when b < f(a), where I did not give the proof) $\int_{f^{-1}(b)}^a\!\!f(x)\,dx > b(a - f^{-1}(b)))$.