Proof of Young's inequality

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.

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:

$\hspace3in$

• 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? Commented Feb 3, 2015 at 18:08
• 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. Commented Dec 21, 2016 at 2:36
• @AOrtiz: fixed. Commented Dec 21, 2016 at 16:09
• Shouldn't the upper bound in the first integral be $f(a)$ instead of $b$? Commented Dec 21, 2016 at 16:15
• @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. Commented Aug 7, 2019 at 18:08

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 $$\int_0^a f(x)\, dx + \int_0^b f^{-1}(x)\, dx = \int_C u\, dv + v\, du = \int_C d(uv) = af(a) = ab.$$

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