A friend has submitted this problem to me:

Let $0<a<b<1$ and $f:[0,1]\to \mathbb R$ be a differentiable function such that $$\displaystyle \frac{\int_0^a f(x) dx}{a(1-a)}+\frac{\int_b^1 f(x) dx}{b(1-b)}=0$$

Prove that $\displaystyle \left| \int_0^1 f(x)dx\right|\leq \left(\frac{b-a}2\right)\sup_{x\in [0,1]}\left|f'(x)+2\normalsize{\frac{\int_0^a f(t) dt}{a(1-a)}}\right|$

I haven't solved this yet, but I've made some progress:

  • Since $\frac{1}{1-a}\left(\frac{1}{a}\int_0^a f(x) dx\right)+ \frac{1}{b}\left(\frac{1}{1-b}\int_b^1 f(x) dx\right)=0$, the Mean Value Theorem yields the existence of $\xi_a\in[0,a]$ and $\xi_b\in[b,1]$ such that $$bf(\xi_a)+(1-a)f(\xi_b)=0$$ This implies $f(\xi_a)$ and $f(\xi_b)$ have opposite signs, thus there is some $c\in [\xi_a,\xi_b]$ such that $f(c)=0$.

  • We may suppose WLOG that $\sup_{x\in [0,1]} |f'(x)|<\infty$ (there's nothing to prove otherwise). Let $M=\sup_{x\in [0,1]} |f'(x)|$.

One can write $$\displaystyle \begin{align}\int_0^1 f(x)dx &= \int_0^a f(x)dx + \int_a^b f(x)dx + \int_b^1 f(x)dx \\ &= \frac{\int_0^a f(x) dx}{a(1-a)} \left( a(1-a) -b(1-b)\right) + \color{red}{\int_a^b f(x)dx} \end{align} $$

Note that $\displaystyle \color{red}{\int_a^b f(x)dx} = \color{green}{\int_a^b (f(x)-f(a))dx} + \color{blue}{(b-a)f(a)}$.

Since $f(a) = f(a)-f(c) = f'(\xi_c) (a-c)$, we have $|f(a)|\leq M |a-c|\leq M$.

The Mean Value Theorem also gives $\displaystyle \left| \int_a^b (f(x)-f(a))dx \right| \leq M \int_a^b (x-a) dx = M \frac{(b-a)^2}2$. Putting everything together we have the estimate $$ \left| \int_0^1 f(x)dx\right|\leq \left(\frac{b-a}2\right) \left[2 \left|\frac{\int_0^a f(x) dx}{a(1-a)}\right| \underbrace{\frac{ \left( a(1-a) -b(1-b)\right)}{b-a}}_{\leq 1} + \color{green}{M (b-a)} +\color{blue}{2M } \right]$$

and we get the bound $$ \left| \int_0^1 f(x)dx\right|\leq \left(\frac{b-a}2\right)\left(2\normalsize{\frac{\int_0^a f(t) dt}{a(1-a)}} + 3M\right) $$

which is not as sharp as what's required...

Note: this problem is similar to this Prove an integral inequality $|\int\limits_0^1f(x)dx|\leq\frac{1-a+b}{4}M$

I've tried to apply similar techniques, to no avail.

  • $\begingroup$ Wouldn't proving that $\sup_{x \in [0, 1]} \displaystyle \left |f'(x) + 2\dfrac{\int\limits_{0}^{a} f(t) dt}{a(1 - a)}\right | \leq 3M + \left|2\dfrac{\int\limits_{0}^{a} f(t) dt}{a(1 - a)}\right|$ suffice? I'm not exactly sure, but I recall that the suprema have a property of $\sup A + B = \sup A + \sup B$. $\endgroup$
    – Raito
    Jan 6 '17 at 18:10

Let $$\lambda = \frac{\int_0^a f(x)\mathrm{d}x}{a(1-a)}= -\frac{\int_b^1 f(x)\mathrm{d}x}{b(1-b)}.$$ Set $g(x) = f(x)+2\lambda x - \lambda$. Notice that $$\int_0^a g(x)\mathrm{d}x = \int_0^a f(x)\mathrm{d} x + \lambda a^2 - \lambda a = \lambda a (1-a) + \lambda a ^2 - \lambda a = 0.$$ Similarly, we obtain $$\int_b^1 g(x)\mathrm{d}x = 0.$$

Now the left hand side of inequality becomes $$\left|\int_0^1 \left(g(x) - 2\lambda x + \lambda\right) \mathrm{d}x\right| = \left|\int_0^1 g(x) \mathrm{d}x \right|,$$ whereas the right hand side becomes $$\frac{b-a}{2}\cdot \sup_{0 \le x\le 1}|g'(x)|.$$

Indeed, we have reformulated the problem in terms of $g$. Note that the top answer for Prove an integral inequality $\left|\int_0^1f(x)dx\right|\le\frac{1−a+b}{4}\sup_{0\le x \le 1}|f'(x)|$ actually proves a stronger statement, and it is exactly what we need.

  • $\begingroup$ Nice solution. How did you notice that adding an affine function to $f$ would change the setup to the one in the other question ? $\endgroup$ Jan 7 '17 at 19:28
  • 1
    $\begingroup$ @LeGrandDODOM It is natural to look at the integral of $f'(x) + 2\lambda$ :) $\endgroup$
    – Zilin J.
    Jan 7 '17 at 19:53

Consider the function $$ g(t) = \int_0^tf(x)dx + \frac{t(1-t)}{b(1-b)}\int_{b}^1f(x)dx.$$ Differentiating twice gives $$g^{\prime\prime}(t) = f^\prime(t) - \frac{2}{b(1-b)}\int_{b}^1f(x)dx = f^\prime(t) + \frac{2}{a(1-a)}\int_{0}^af(x)dx .$$ The required inequality is then equivalent to proving that $$|g(b)| \leq \frac{b-a}{2}\sup_{x\in[0,1]}|g^{\prime\prime}(x)|. $$ However, since $g(a)=g(1)=0$ we can do this. Clearly we have that $$|g(b)| \leq (b-a) \sup_{x\in[a,b]}|g^\prime(x)|$$ from mean value theorem and so all we need to do now is prove that $$\sup_{x\in[a,b]}|g^\prime(x)| \leq \frac{1}{2}\sup_{x\in[a,b]}|g^{\prime\prime}(x)|.$$ Define another function $h(t)=g((1-a)t+a)$, this is twice differentiable with $h(0)=h(1)=0.$ For each point $x\in(0,1)$, using Taylor's theorem twice, there exists $c_x\in(0,x)$ and $d_x\in (x,1)$ such that $$h^\prime(t) = h^{\prime\prime}(c_x)\frac{(c_x-x)^2}{2} -h^{\prime\prime}(d_x)\frac{(d_x-x)^2}{2} $$ Therefore, $$ |h^\prime(t)| \leq \sup_{x\in[0,1]}|h^{\prime\prime}(x)|\left(\frac{t^2}{2} + \frac{(1-t)^2}{2}\right) \leq \frac{1}{4}\sup_{x\in[0,1]}|h^{\prime\prime}(x)|.$$ Hence we have $$ |g^\prime(t)|\leq \frac{1-a}{4}\sup_{x\in[a,1]}|g^{\prime\prime}(x)|.$$ This gives us our required inequality.


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