8
$\begingroup$

Let $f$ be a twice differentiable function on $\left[0,1\right]$ satisfying $f\left(0\right)=f\left(1\right)=0$. Additionally $\left|f''\left(x\right)\right|\leq1$ in $\left(0,1\right)$. Prove that $$\left|f'\left(x\right)\right|\le\frac{1}{2},\quad\forall x\in\left[0,1\right]$$

The hint we were given is to expand into a first order Taylor polynomial at the minimum of $f'$. So I tried doing that:

As $f'$ is differentiable, it is continuous, and attains a minimum at $\left[0,1\right]$. Thus we can denote $x_{0}$ as the minimum, and expending into a Taylor polynomial of the first order around it gives us, for some $c$ between $x$ and $x_0$.

$$T_{x_{0}}\left(x\right)=f\left(x_{0}\right)+f'\left(x_{0}\right)\left(x-x_{0}\right)+\frac{f''\left(c\right)}{2}\left(x-x_{0}\right)^{2}$$

Now at $x=0$ we have $$T_{x_{0}}\left(0\right)=f\left(x_{0}\right)-x_{0}f'\left(x_{0}\right)+x_{0}^{2}\frac{f''\left(c\right)}{2}=0$$

And at $x=1$ we have $$T_{x_{0}}\left(0\right)=f\left(x_{0}\right)+\left(1-x_{0}\right)f'\left(x_{0}\right)+\left(1-x_{0}\right)^{2}\frac{f''\left(c\right)}{2}=0$$

and I'm pretty much stuck here..

So I tried a different approach using Mean Value Theorem directly, by Rolle's I know the derivative is $0$ somewhere (as $f(0)=f(1)$), suppose at $x_0$, so by mean value theorem $\frac{f'(x)}{x-x_0} =f''(c)\leq1$ for some $c$ and $f'(x)<x-x_0$.

But using this approach as well I'm not sure how to proceed, as this gives me the desired propety only in $1/2$ environment around $x_0$...

Any help?

$\endgroup$
  • 1
    $\begingroup$ Just a brief word of warning. When you write $\min_{x\in [0,1]} g(x)$, this ordinarily means the minimum value of $g$ on the interval, not the minimum point $x_0$. Also: Shouldn't we be talking about the maximum of $|f'|$, not the minimum? $\endgroup$ – Ted Shifrin Jun 9 '15 at 14:47
  • $\begingroup$ @TedShifrin That's a fair point, changed it. About the maximum, the hint I was given was to use the minimum, when I wasn't able to use that I tried thinking of the maximum, but even if I do, everything else is pretty much the same, and in the given expansion I'm not sure how to prove it's smaller than $\frac{1}{2}$ $\endgroup$ – Nescio Jun 9 '15 at 14:53
  • $\begingroup$ OK, I'm thinking. In particular, $f$ has (without loss of generality) a maximum $x_0$ and this will be a point where $f'(x_0)=0$, of course. $\endgroup$ – Ted Shifrin Jun 9 '15 at 14:55
  • 2
    $\begingroup$ A nice source of intuition is to find where the inequality is tight. Often in these kinds of problems this happens for a polynomial. In this case it happens for $f(x)=(x^2-x)/2$ (or with the opposite sign). $\endgroup$ – Ian Jun 9 '15 at 16:35
5
$\begingroup$

You were on the right track, but, as I suggested, the hint should have been to expand about the point $x_0$ where $|f'(x_0)|$ is a maximum.

Fix any $x_0\in [0,1]$ and, using Taylor's Theorem, write $$f(x)=f(x_0) + f'(x_0)(x-x_0) + \frac12 f''(c)(x-x_0)^2\quad\text{for some $c$ between $x_0$ and $x$.}$$ Plugging in $x=0$ and $x=1$ respectively, we arrive at $$f'(x_0) = \frac12\big(f''(c_1)x_0^2 - f''(c_2)(1-x_0)^2\big) \quad\text{for some $c_1$ and $c_2$}.$$ Therefore $|f'(x_0)|\le \dfrac12\big(x_0^2 + (1-x_0)^2\big) \le \dfrac12$, since $x_0\in [0,1]$ is arbitrary.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

This is an equivalent statement:

Let $g:[0,1] \to \mathbb R$ be a continuous function on $[0,1]$. Assume that $g$ is differentiable on $(0,1)$,
$$ \int_0^1 g(x) dx = 0 \qquad \text{and} \qquad \forall x \in (0,1) \quad \bigl| g'(x) \bigr| \leq 1. \tag{1}\label{1} $$ Then $|g(x)| \leq 1/2$ for all $x \in [0,1]$.

Proof. By the Mean value theorem, for all $x,y \in [0,1]$ such that $x \neq y$ there exists $c \in (0,1)$ such that $$ \frac{g(y) - g(x)}{y-x} = g'(c). $$ By the second assumption in $\eqref{1}$, for all $x,y \in [0,1]$ such that $x \neq y\ $ we have $$ \left| \frac{g(y) - g(x)}{y-x} \right| \leq 1. $$ Consequently, for all $x,y \in [0,1]$ we have $$ -|y-x| \leq g(y) - g(x) \leq |y-x|. $$ Integrating the last inequality with respect to $x$ over $[0,1]$ and using the first assumption in $\eqref{1}$, we get $$ -\int_0^1 |y-x| dx \leq g(y) \leq \int_0^1 |y-x| dx. $$ Thus, for all $y \in [0,1]$ we have $$ \bigl| g(y) \bigr| \leq \int_0^1\!\! |y-x| dx =\frac{1}{2}\bigl( y^2 + (1-y)^2 \bigr) = \Bigl(y-\frac{1}{2}\Bigr)^2 + \frac{1}{4} \leq \frac{1}{2}. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.