real-analysisriemann-integrationyoung-inequality
Suppose that f is continuous increasing function with $f(0)=0$ then prove that for $a,b>0$ we have Young's inequality
$$ab\leq \int_{0}^{a} f(x)dx+\int_{0}^{b}f^{-1}(x)dx$$
My attempt is incorrect: I didn't argue using lower and upper sums which i assume you need to use, any help would be appreciated thanks.
First note that we have $$ab \leq \int_0^{a} f(x) dx + \int_0^{b} f^{-1}(x) dx $$ for any strictly increasing integrable function $f(x)$. The geometric interpretation is from looking at the area of the rectangle with coordinates $(0,0)$,$(a,0)$,$(a,b)$ and $(0,b)$ and comparing it with the areas given by the integrals. From the image it is also clear that the equality hold only when $b=f(a)$.
To get the Young's inequality, choose $f(x) = x^{p-1}$.
I have added the following picture for clarity.
The image was made using grapher and some post processing was done using LaTeXiT and preview on Mac OSX.
Assuming $f(a)>b$, we have: $$ \color{red}{\int_{0}^{a} f(x)\,dx} = \mu\left(\{(x,y)\in[0,a]\times[0,f(a)]: 0\leq y\leq f(x)\}\right) $$ and: $$ \color{blue}{\int_{0}^{b} f^{-1}(y)\,dy} = \mu\left(\{(x,y)\in[0,a]\times[0,b]: 0\leq x\leq f^{-1}(y)\}\right) $$ so the sum of the two integrals surely exceeds $\mu\left([0,a]\times[0,b]\right)=ab$.
By the way, it is a lot easier just to draw a picture:
$\hspace3in$
Best Answer
We can make use of the Riemann-Stieltjes integral. Proceeding we enforce the substitution $x=f(t)$ to obtain
$$\begin{align} \int_0^b f^{-1}(x)\,dx&=\int_{0}^{f^{-1}(b)} t\,df(t)\\\\ &=bf^{-1}(b)-\int_0^{f^{-1}(b)}f(t)\,dt\tag 1 \end{align}$$
Using $(1)$ we see that
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx=\int_{f^{-1}(b)}^af(x)\,dx+bf^{-1}(b)\tag 2$$
If $a>f^{-1}(b)$, then $\int_{f^{-1}(b)}^af(x)\,dx\ge b(a-f^{-1}(b))$ since $f$ is increasing. Therefore, we assert that
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 3$$
If $a<f^{-1}(b)$, then $\int_{f^{-1}(b)}^af(x)\,dx=-\int_a^{f^{-1}(b)}f(x)\,dx\ge -b(f^{-1}(b)-a)$ since $f$ is increasing. Therefore, we assert that
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 4$$
Finally, if $a=f^{-1}(b)$, then we see directly from $(2)$ that
$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 5$$
Putting together $(3)$, $(4)$, and $(5)$ yields the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab}$$
as was to be shown!