Suppose that $f\colon [0,\infty)\to \mathbb{R}$ is a continuous strictly increasing function with $f(0)=0$. Prove that for $a,b \gt 0$ we have Young's inequality
$$ ab \le \int_0^af(x)dx+\int_0^bf^{-1}(x)dx$$ and that equality holds if and only if $b=f(a)$.
In this post the following proof is given by kobe:
Let $C$ be the graph of $v = f(u)$ over the interval $[0,f^{-1}(b)]$. If $f(a) > b$, then $f^{-1}(b) < a$, in which case
\begin{align}\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx &= \int_0^{f^{-1}(b)} f(x)\, dx + \int_0^b f^{-1}(x)\, dx + \int_{f^{-1}(b)}^a f(x)\, dx\\ &= \int_C u\, dv + v\, du + \int_{f^{-1}(b)}^a f(x)\, dx\\ &= \int_C d(uv) + \int_{f^{-1}(b)}^a f(x)\, dx\\ &= bf^{-1}(b) + \int_{f^{-1}(b)}^a f(x)\, dx\\ &> bf^{-1}(b) + b(a - f^{-1}(b))\\ &= ab \end{align}
Similarly if $f(a) < b$, then $\int_0^a f(x)\, dx + \int_0^b f(x)\, dx > ab$. If $f(a) = b$, then $$\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx = \int_C u\, dv + v\, du = \int_C d(uv) = af(a) = ab.$$
I would like to focus on the following equality:
$$\int_0^{f^{-1}(b)} f(x)\; dx=\int_C v\;du$$
We parameterize $C$ using the curve $$\gamma(x)=(u(x), v(x)):=(x, f(x)), \quad x\in [0, f^{-1}(b)]$$ then for definition of line integral we have that $$\int_0^{f^{-1}(b)}f(x)\; dx=\int_\gamma v\; du.$$
Question 1. Is $\gamma$ piecwise regular or regular?
For me the answers is yes, because the function $f$ is monotone and then $f'(x)$ exists finite for almost every $x\in [0, f^{-1}(b)]$.
Thus exists a null set on which $f'$ may be not exists.
Question 2 Do the hypotheses we have allow us to say something more about this set of zero measure, such as that it is finite?
Thanks!
