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The following problem is from Spivak's Calculus.

Suppose that $f$ is a continuous 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 $b=f(a)$.

It is enough to consider the case $f(a) \gt b$, and show that the strict inequality occurs in this case.

I've tried proving this using the theorem $$ \int_a^bf^{-1}=bf^{-1}(b)-af^{-1}(a)-\int_{f^{-1}(a)}^{f^{-1}(b)}f$$

but I got stuck along the way.

How may I show this rigorously using the definition or properties of integrals? Any hint, suggestions or solutions would be appreciated.

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2 Answers 2

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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:

$\hspace3in$enter image description here

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  • $\begingroup$ Why can the sum not exceed ab? And I can clearly see it from the picture, but I was wondering how I can prove this without resorting to a diagram. How can I guarantee the fact that the result on the diagram must happen? $\endgroup$ Commented Feb 3, 2015 at 18:08
  • $\begingroup$ Your conclusion that the sum of the integrals cannot exceed $ab$ seems to conflict with the claim that $ab$ is $\le$ than the sum of the integrals. $\endgroup$
    – Alex Ortiz
    Commented Dec 21, 2016 at 2:36
  • $\begingroup$ @AOrtiz: fixed. $\endgroup$ Commented Dec 21, 2016 at 16:09
  • $\begingroup$ Shouldn't the upper bound in the first integral be $f(a)$ instead of $b$? $\endgroup$
    – Dominik
    Commented Dec 21, 2016 at 16:15
  • $\begingroup$ @nomadicmathematician $\mu$ here is used as a measure; in this context, $\mu(A)$ simply means the area of set $A$. All measures enjoy the property $\mu(A \cup B) = \mu(A) + \mu(B)$ for disjoint sets $A,B$ used here, and you can see this agrees with intuition for areas. $f(a)>b$ implies the first integral is greater than $\mu(\{(x,y): x\in[0,a],y\in[0,b], 0\leq y \leq f(x)\})$. Now observe that for every $(x,y) \in [0,a]\times [0,b]$, either $y \leq f(x)$ or $x \leq f^{-1}(y)$. There is a slight technicality; the sets intersect on $y=f(x)$, but this set has $0$ area/measure, and can be dropped. $\endgroup$ Commented Aug 7, 2019 at 18:08
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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 $$\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx = \int_C u\, dv + v\, du = \int_C d(uv) = af(a) = ab.$$

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  • $\begingroup$ Could you explain to me how the second equality holds ($\int_0^{f^{-1}(b)} f(x)\, dx + \int_0^b f^{-1}(x)\, dx + \int_{f^{-1}(b)}^a f(x)\, dx = \int_C u\, dv + v\, du + \int_{f^{-1}(b)}^a f(x)\, dx$) ? $\endgroup$
    – John Mars
    Commented Dec 21, 2020 at 17:14
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    $\begingroup$ @JohnMars the integral $\int_0^{f^{-1}(b)} f(x)\, dx = \int_C v\, du$ and $\int_0^b f^{-1}(x)\, dx = \int_C u\, dv$, so $\int_0^{f^{-1}(b)} f(x)\, dx + \int_0^b f^{-1}(x)\, dx = \int_C u\, dv + v\, du$. $\endgroup$
    – kobe
    Commented Dec 23, 2020 at 19:36

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