5

Let $f:[0,1]\to \mathbb{R}$ be a differentiable function such that $\displaystyle \int \limits _0^1f(t)\,dt=1$, $f(0)=0$, $f(1)=0$. Prove that there exists an $x_0\in (0,1)$ such $|f'(x_0)|\geq 4$.

I'm trying to use mean value theorem on this but its leading to a result I already know that the there would exist a point where tangent is $0$.

q123LsaB
  • 327
  • 1
  • 12
  • edited for clarification. thanks for pointing out. – q123LsaB Aug 06 '22 at 13:19
  • With humility, the conditions $f(0) = 0$ and $f(1) = 0$ seem to imply (by Rolle's theorem) that $f^{ \prime} (x_0) \ge 4$ for all $x_0 \in (0,1)$ is impossible. It appears that no one could prove that, regardless of how the question is phrased. – Jeffrey Harkness Aug 06 '22 at 15:05
  • @JeffreyHarkness why is that? – q123LsaB Aug 06 '22 at 16:58
  • Rolle's theorem guarantees that for a continuous function on the interval $(a,b)$ with $f(a) = f(b) = 0$, then $\exists x_0 \in (a,b)$ such that $f ^{\prime} (x_0) = 0$. Then $f ^{\prime} (x_0) \ge 4$ is not true at this point. Another theorem states that if $f$ is differentiable at a point, then it is continuous, so the continuity condition is met. See https://en.wikipedia.org/wiki/Rolle's_theorem and https://en.wikipedia.org/wiki/Continuous_function#Relation_to_differentiability_and_integrability – Jeffrey Harkness Aug 06 '22 at 18:35
  • yes but it doesn't say that it needs to hold true for all $x_o$, I just need to show that there would exist at least 1 point satisfying the condition.. – q123LsaB Aug 06 '22 at 18:38
  • I was responding to @Surb 's comment regarding lack of clarity on whether $f ^{\prime} (x_0) \ge 4$ for some $x_0 \in (0,1)$ or for all $x_0 \in (0,1)$. I was merely illustrating that $f ^{\prime} (x_0) \ge 4$ appeared to be impossible for all $x_0 \in (0,1)$ based on the original problem statement. – Jeffrey Harkness Aug 06 '22 at 22:51

3 Answers3

3

The fundamental theorem of calculus is usually a good thing to try: if to the contrary $|f'(t)| < 4$ for all $t$ you would have $$f(x) = f(x) - f(0) = \int_0^x f'(t) \, dt < 4x$$ for all $x \in [0,1]$. In particular, $$\int_0^{1/2} f(x) \, dx < \int_0^{1/2} 4x \, dx = \frac 12.$$

This is only a partial solution, but it completely neglected the hypothesis that $f(1) = 0$. What will that imply?

Umberto P.
  • 52,165
2

the function $$f(x) = \begin{cases} 3x \hspace{1cm} & x \leq \frac{11}{15} \\ 3x - \frac{6075}{32}(x-\frac{11}{15})^2 + \frac{70875}{128}(x-\frac{11}{15})^3 \hspace{0.5cm} &x \geq \frac{11}{15} \end{cases}$$

is such that

(i) $f \in C^1(\mathbb{R})$

(ii) $ f'(x) < 4 \;\; \forall x \in [0,1]$

(iii) $f(0) = 0$, $f(1) = 0\;\;\;, \;\;\;\int_0^1{f(x)dx} = 1$

Therefore what you're trying to prove is false.

Maybe if you substitute $4$ with a smaller number the statement becomes true.

To prove that (i),(ii) and (iii) holds I recommend to use a graphic calculator like wolfram alpha or matlab and to not prove it analytically, although you can do it if you wish but it is very tedious

Paul
  • 1,314
  • 1
    I think that, in order for the statement to hold, you need to substitute 2 instead of 4. – Paul Aug 06 '22 at 15:09
  • I agree; the statement as written is not correct. Another possible fix might be that we must have $|f'(x_0)| \ge 4$ for some $x_0 \in [0,1]$. This seems to give the correct result considering Umberto's answer. By the way, how did you come up with the specific example you gave? – legionwhale Aug 06 '22 at 15:53
  • The idea is that if you let $f(x) = mx$ on a certain set of the form [0,a], where 0 < a < 1 and $2 < m < 4$, then you can find a certain a and a a certain function $g : [a,1] -> \mathbb{R}$ such that the function $f(x) = mx$ for $0 \leq x \leq a$ and $f(x) = g(x)$ for $a \leq x \leq 1$ satisfies all the conditions but $f'(x) \leq m$ for any $x \in [0,1]$ In particolar I let $m = 3$ and used some kind of numerical analysis interpolation technique to find the exact expression of $g$ – Paul Aug 06 '22 at 16:44
  • Of course. But my question is how did you find such a $g$? – legionwhale Aug 06 '22 at 16:45
  • 2
    I let $m = 3$ , $a = \frac{11}{15}$, $g(x) = mx + A(x - a)^2 + B(x - a)^3$ and found $A$ and $B$ such that the condition holds. The fact that $g'(x) < 4 ;; \forall x \in [a,1]$ was just luck – Paul Aug 06 '22 at 16:49
  • I have edited the question, thank you for your clarification. – q123LsaB Aug 06 '22 at 17:03
0

As has been shown in the other answer this statement seems to be false. However if you have $\left|f'\left(x_0\right)\right| \ge 4$ instead of $f'(x_0) \ge 4$ and $x_0 \in[0,1]$. To prove that let $$g(x) = \int_0^x\left(f(t) + f(1-t)\right)\mathrm d t$$

\begin{align} \frac 14\sup\limits_{t\in [0,1]} \left|f'(t)\right| &\ge \frac 14\times \frac12\sup\limits_{t\in \left[0,\frac 12\right]} \left|f'(t) - f'(1-t)\right|\\ &= \frac 12 \sup_{t\in \left[0,\frac12\right]} \left|g''(t)\right|\left(\frac 12\right)^2\\ & \ge \left|g\left(\frac12\right)-g(0)-\frac12 g'(0)\right| = 1 \end{align}

and you have what you are looking for using the continuity of $f'$.

Kroki
  • 13,135
  • sorry, I do not understand your solution. I don't know what "sup" is, I'm in highschool and have just started applying derivatives and the mean value theorems, with some idea with integral calculus. – q123LsaB Aug 06 '22 at 17:15