I was practicing calculus today and stumbled across this problem. I have tried solve using properties of inequalities but it doesn't get me far.

Let $f:[0,1] \to \mathbb{R}$ be continuously differentiable function and $$ \bigg|\space\frac{f(0)+f(1)}{2} -\int_0^1 f(x)\space dx \space\bigg|\le \alpha \space\space \max( |f'(x)|), \space\space x\in[0,1]$$

Then find the possible values of $\space\alpha$ .

Any ideas that can solve this easily ? Thank you.

  • 1
    $\begingroup$ To denote set inclusion, use $\in$, not $\epsilon$. Code is \in. $\endgroup$
    – K.defaoite
    Jun 3, 2020 at 15:09
  • $\begingroup$ You can assume without loss of generality $f(0)=0$, and as long as $f$ is not the zero function, you can assume $\max|f(x)|=1$. $\endgroup$
    – M. Nestor
    Jun 3, 2020 at 15:33

1 Answer 1


\begin{align} \bigg|\,\frac{f(0)+f(1)}{2}-\int_0^1f(x)dx\,\bigg|=&\\ \bigg|\,\frac{f(0)+f(1)}{2}-\int_0^1\left(f(0)+\int_0^xf'(y)dy\right)dx\,\bigg|=&\\ \bigg|\,\frac{f(1)-f(0)}{2}-\int_0^1\int_0^xf'(y)dydx\,\bigg|=&\\ \bigg|\,\frac{1}{2}\int_0^1f'(y)dy-\int_0^1\int_0^xf'(y)dydx\,\bigg|= &\\ \bigg|\,\frac{1}{2}\int_0^1f'(y)dy-\int_0^1\int_y^1f'(y)dxdy\,\bigg|= &\\ \bigg|\,\int_0^1f'(y)\left(\frac{1}{2}-\int_y^1dx\right)dy\,\bigg|= &\\ \bigg|\,\int_0^1\ \left(y-\frac{1}{2}\right)f'(y)dy\,\bigg|\le&\\ \int_0^1\ \bigg|\,\left(y-\frac{1}{2}\right)f'(y) \bigg|\, dy\le&\\ \max_{0\le x\le1}(|f'(x)|)\int_0^1\bigg|y-\frac{1}{2}\bigg|dy=&\\ \max_{0\le x\le1}(|f'(x)|)\left(\int_0^\frac{1}{2}\left(\frac{1}{2}-y\right) dy+ \int_\frac{1}{2}^1\left(y-\frac{1}{2}\right) dy\right)\\ =\frac{1}{4} \max_{0\le x\le1}(|f'(x)|) \end{align}

  • The first, third and sixth equations follow from the second fundamental theorem of calculus: $$ f(x)=f(0)+\int_0^xf'(y)dy\\ f(1)-f(0)=\int_0^1f'(y)dy\\ \int_y^1dx=1-y\ . $$

  • The second and fifth equations follow from algebraic manipulations, $\displaystyle\ \int_0^1f(0)\,dx=f(0) \ $, and the linearity of integrals.

  • The fourth equation follows from the fact that a double integral $\displaystyle\ \iint_R\varphi(x,y)dxdy\ $ over the set $\ R=\left\{\left.(x,y)\in\mathbb{R}^2\right|0\le y\le x\le1\right\}\ $ can be evaluated either as $\displaystyle\ \int_0^1\int_0^x \varphi(x,y)\,dydx\ $ or as $\displaystyle\ \int_0^1\int_y^1 \varphi(x,y)\,dxdy\ $.
  • The first inequality is an instance of the general theorem that $\displaystyle\ \bigg|\int_a^bg(y)\,dy\bigg|\le\int_a^b\bigg| g(y) \bigg|\,dy\ $ and the second is an instance of the general theorem that if $\ 0\le\varphi(x)\ $ and $\ 0\le\psi(x)\le M\ $, then $\ \int_a^b\varphi(y)\psi(y)\,dy\le$$M\int_a^b\varphi(y)\,dy\ $.
  • The last two equations are just steps in the evaluation of the integral $\displaystyle\ \int_0^1\bigg|y-\frac{1}{2}\bigg|dy\ $, which is done by splitting its range into the intervals $\ \left[0,\frac{1}{2}\right]\ $ where $\ y\le \frac{1}{2}\ $ and hence $\ \left|y-\frac{1}{2}\right|= \frac{1}{2}-y\ $, and $\ \left[\frac{1}{2},1\right]\ $ where $\ y\ge \frac{1}{2}\ $ and hence $\ \big|y-\frac{1}{2}\big|=y-\frac{1}{2}\ $.

It follows from all this that $\ \alpha\ $ could be any real number that is at least as large as $\ \frac{1}{4}\ $. You can show that $\ \alpha\ $ cannot be less than $\ \frac{1}{4}\ $ by taking $\ f\ $ to be given by $$ f(y)=\cases{\frac{1}{2}-y& for $\ 0\le y\le \frac{1}{2}-\epsilon\ $,\\ \frac{\left(y-\frac{1}{2}\right)^2}{2\epsilon} & for $\ \frac{1}{2}-\epsilon\le y\le \frac{1}{2}+\epsilon\ $,\\ y-\frac{1}{2}& for $\ \frac{1}{2}-\epsilon\le y\le 1\ $, } $$ where $\ 0<\epsilon<\frac{1}{2}\ $. This $\ f\ $ is continuously differentiable on $\ [0,1]\ $, with $\displaystyle\ \max_{0\le x\le1}(|f'(x)|)=1\ $ and $\displaystyle\ \frac{f(0)+f(1)}{2}-$$\displaystyle\int_0^1f(x)dx=$$\displaystyle\frac{1}{4}-\frac{2\epsilon^2}{3}\ $, and this can be made as close as you please to $\ \frac{1}{4}\ $ by making $\ \epsilon\ $ sufficiently close to $\ 0\ $.

  • $\begingroup$ @Ionzaleggiera max(|f′(x)|) not max(|f(x)|) $\endgroup$ Jun 4, 2020 at 11:58
  • $\begingroup$ @Anand Maneesh I have updated my answer in response to the change in the question. Please note that I have interpreted the question as requiring the inequality to hold for all continuously differentiable $\ f\ $. Without that requirement the question wouldn't seem to me to make a lot of sense. $\endgroup$ Jun 5, 2020 at 4:42
  • $\begingroup$ Hey There! I did not understand anything from step two.I am only equipped with baby calculus. $\endgroup$ Jun 9, 2020 at 16:57
  • 1
    $\begingroup$ @Anand Maneesh I have added extra details of why each equation and inequality is true. They all follow from fairly elementary results of calculus, although I'm not sure they would qualify as "baby calculus". The least elementary bits are the double integral identity and the generalised triangle inequality, $\displaystyle\ \bigg|\int_a^bg(y)\,dy\bigg|\le\int_a^b\bigg| g(y) \bigg|\,dy\ $. I can't see any way of answering the question without using those results. $\endgroup$ Jun 11, 2020 at 14:56
  • $\begingroup$ Thank you very much,The solution is beautiful. $\endgroup$ Jun 12, 2020 at 3:27

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