# $f,g$ are nonnegative decreasing and $\int_0^x f \le \int_0^x g$ for each $x \in [0,1]$ then $\int_0^1 f^2 \le \int_0^1 g^2$

Here's what I am trying to prove:

Let $$f,g :[0,1] \to \mathbb R$$ be nonnegative decreasing functions and suppose that $$\int_0^x f \le \int_0^x g$$ holds for each $$x \in [0,1]$$ then show that $$\int_0^1 f^2 \le \int_0^1 g^2$$.

I tried to prove it but I think there's possibly some errors. Here's my attempt, nonetheless:

Let $$\alpha =\sup \{ x\in [0,1] : g(x). Now, if $$y>\alpha$$ then $$g(y)\ge f(y)$$. Since $$g$$ and $$f$$ are decreasing, we have $$(g-f)(g+f)\ge (g(1)+f(1) )(g-f)$$ on $$(\alpha,1]$$ and hence $$\int_\alpha ^1 (g^2-f^2) \ge0$$ since $$g-f \ge 0$$ on $$(\alpha,1].$$

Since $$g$$ and $$f$$ are decreasing, we have $$(g-f)(g+f)\ge (g(1)+f(1) )(g-f)$$ on $$[0,\alpha)$$ and hence $$\int_0 ^\alpha (g^2-f^2) \ge0$$ since $$\int_0^\alpha (g-f) \ge 0$$ by assumption.

And thus, the result follows.

Is my attempt correct? Also alternative proofs are welcome!

• It seems that you are assuming that $g(1)+f(1) \ge 0$. Nov 20, 2022 at 18:24
• Oops, yes, I missed that. – But why is $(g-f)(g+f)\ge (g(1)+f(1) )(g-f)$ on $[0, \alpha)$? Nov 20, 2022 at 18:44
• Yes, same question: on $[0,\alpha),$ $g-f$ might be sometimes $<0.$ Nov 20, 2022 at 18:45
• Possibly helpful: math.stackexchange.com/q/1041303/42969 Nov 20, 2022 at 18:49
• @AnneBauval ah yes. There's a mistake. Nov 20, 2022 at 19:15

As pointed out in the comments, your proof is not correct because $$(g-f)(g+f)\ge (g(1)+f(1) )(g-f)$$ does not necessarily hold for all $$x \in [0, \alpha)$$. Also I cannot see where the condition $$\int_0^x f \le \int_0^x g$$ for all $$x \in [0, 1]$$ is used.

For a proof we can use the “Second mean value theorem for definite integrals” in the following form:

Let $$h: [a, b] \to \Bbb R$$ be a nonnegative, decreasing function, and $$\phi: [a, b] \to \Bbb R$$ be an integrable function. Then $$\int_a^b h(x) \phi(x) \, dx = h(a+) \int_a^c \phi(x) \, dx$$ for some $$c \in [a, b]$$.

Here $$h(a+)$$ denotes the one-sided limit $$\lim_{x \to a+} h(x)$$.

This can be applied (twice) to our problem: There are $$c, d \in [0, 1]$$ such that \begin{align} \int_0^1 \bigl(g(x)^2-f(x)^2\bigr)\, dx &= \int_0^1 g(x)\bigl(g(x)-f(x)\bigr) \, dx + \int_0^1 f(x)\bigl((g(x)-f(x)\bigr) \, dx \\ &= g(0+) \int_0^c \bigl(g(x)-f(x)\bigr) \, dx + f(0+) \int_0^d \bigl(g(x)-f(x)\bigr) \, dx \\ &\ge 0 \, . \end{align}

Remark: A proof of the second mean value theorem for Lebesgue-integrable functions can be found e.g. in Hobson, E. W. (1909). On the Second Mean-Value Theorem of the Integral Calculus. https://doi.org/10.1112/plms/s2-7.1.14 .

Let $$F(x) = \int_0^x f(t) \, dt$$, and $$G(x) = \int_0^x g(t) \, dt$$. Integrating by parts we obtain $$\int_0^1 f(t)^2 \, dt = F(1)f(1) - \int_0^1 F(t) \, df(t) .$$ Since $$df$$ is a negative measure, it follows that this is less than or equal to $$G(1)f(1) - \int_0^1 G(t) \, df(t) = \int_0^1 f(t) g(t) \, dt .$$ Therefore by Cauchy-Schwarz, we have $$\int_0^1 f^2 \le \int_0^1 fg \le \left(\int_0^1 f^2\right)^{1/2} \left(\int_0^1 g^2\right)^{1/2} .$$

• Instead of CS you can also do IBP twice: $\int_0^1 f(t)^2 \, dt \le \int_0^1 f(t)g(t) \, dt \le \int_0^1 g(t)^2 \, dt$. Dec 4, 2022 at 19:23

Lemma: $$a_1\ge a_2\ge\cdots\ge a_n\ge0$$ and $$b_1\ge b_2\ge\cdots\ge b_n\ge0$$. $$\sum_{i=1}^ka_i\le\sum_{i=1}^kb_i$$ for all $$k$$. Then $$\sum_{i=1}^na_i^2\le\sum_{i=1}^nb_i^2$$.
Proof: Let $$\ell_k= \sum_{i=1}^ka_i-\sum_{i=1}^kb_i$$. The condition says $$\ell_k\le0$$ for all $$1\le k\le n$$. By definition, $$\ell_0=0$$. Note that $$a_k-b_k=\ell_k-\ell_{k-1}$$ for all $$1\le k\le n$$. \begin{aligned} \sum_{i=1}^na_i^2-\sum_{i=1}^nb_i^2 &=\sum_{i=1}^n(a_i+b_i)(a_i-b_i)\\ &=\sum_{i=1}^n(a_i+b_i)(\ell_i-\ell_{i-1})\\ &=(a_n+b_n)\ell_n-(a_1+b_1)\ell_0 + \sum_{i=1}^{n-1}((a_i-a_{i+1})+(b_i-b_{i+1}))\ell_i\\ &\le0 \end{aligned}

The lemma is a slight variant of Karamata's inequality with the convex function $$f=x^2$$.

The proposition in the question is the continuous variant of the lemma.

Fix a positive integer $$n$$. Let $$\ a_i=\int_{\frac{i-1}n}^\frac in f$$, $$\ b_i=\int_{\frac{i-1}n}^\frac in g$$ for all $$1\le i\le n$$.

• $$f$$ and $$g$$ are nonnegative decreasing implies $$a_1\ge a_2\ge\cdots\ge a_n\ge0$$ and $$b_1\ge b_2\ge\cdots\ge b_n\ge0$$.
• $$\int_0^x f \le \int_0^x g$$ holds for each $$x \in [0,1]$$ implies $$\sum_{i=1}^ka_i=\int_0^\frac kn f \le \int_0^\frac kn g=\sum_{i=1}^kb_i$$ for all $$k$$.

The lemma tells that $$\sum_{i=1}^na_i^2\le\sum_{i=1}^nb_i^2$$. Multiplying by $$n$$, we have $$\sum_{i=1}^nna_i^2\le\sum_{i=1}^nnb_i^2.$$ Letting $$n\to\infty$$, we obtain $$\int_0^1f^2\le\int_0^1g^2$$ since $$\lim_{n\to\infty}\sum_{i=1}^nna_i^2=\int_0^1f^2\quad \text{and}\quad\lim_{n\to\infty}\sum_{i=1}^nnb_i^2=\int_0^1g^2,$$ which are easy to prove because $$f$$ and $$g$$ are monotone.

Here is a minor generalization, which can be proved similarly.

Let $$f,g :[0,1] \to \mathbb R$$ be nonnegative decreasing functions and suppose that $$\int_0^x f \le \int_0^x g$$ holds for each $$x \in [0,1]$$. Then we have $$\int_0^1 (\phi\circ f) \le \int_0^1 (\phi\circ g)$$, where $$\phi$$ is a nonnegative increasing convex function on $$\Bbb R_{\ge0}$$.