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I'm working through the Spivak Calculus book, and am having some difficulty with the following problem. Assume that $f$ is a second differentiable function, $f(0) = 0$, $f(1) = 1$, and $f'(0) = f'(1) = 0$. Prove that $\exists c\in [0,1]$ such that $|f''(c)|\geq 4$. The hint that the text provides is prove that either $f''(x) > 4$ for some $x\in [0, \frac{1}{2}]$ or $f''(x) < -4$ for some $x\in [\frac{1}{2}, 1]$.

Using the Mean Value theorem, it's easy enough to prove that that $\exists c_1\in (0,1)$ and $f'(c_1) = \frac{f(1) - f(0)}{1-0} = 1$.

From here, you can apply MVT to $f'$ on $[0,c_1]$ and $[c_1, 1]$ giving

$\exists c_2\in (0,c_1)$ where $f''(c_2) = \frac{1}{c_1}$

and

$\exists c_3\in (c_1, 1)$ where $f''(c_3) = \frac{-1}{1-c_1}$.

From here, using the assumption that $c_1 \leq \frac{1}{4}$ yields $f''(c_2) \geq 4$ or $c_1 \geq \frac{3}{4}$ yields $f''(c_3) \leq -4$.

However, I am unsure of what to do if $\frac{1}{4} < c_1 < \frac{3}{4}$. I have been told that Cauchy's Mean Value Theorem (sometimes called the Generalized Mean Value Theorem) is the way to go concerning this problem, but I'm unable to make it work. Any thoughts would be greatly appreciated.

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marked as duplicate by Paramanand Singh calculus Jun 22 '17 at 5:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I edited my answer below to include a proof using CMVT. $\endgroup$ – zhw. Jun 21 '17 at 18:27
  • $\begingroup$ This is appearing too frequently and there is a list of duplicates already available. $\endgroup$ – Paramanand Singh Jun 22 '17 at 5:01
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You're right, Cauchy's MVT is a nice way to go: There exists $a\in (0,1)$ such that $f(a) = 1/2.$ Apply CMVT twice to see

$$\frac{1}{2a^2}=\frac{f(a)}{a^2} = \frac{f(a)-f(0)}{a^2-0^2} = \frac{f'(c)}{2c} = \frac{f'(c)-f'(0)}{2c-2\cdot 0}=\frac{f''(d)}{2}. $$

This implies $|f''(d)| \ge 1/a^2.$ A similar argument, using $f(1) = 1, f'(1) = 0,$ gives $d'$ such that $|f''(d')| \ge 1/(1-a)^2.$

Now one of $1/a^2, 1/(1-a)^2$ has to be at least $4,$ giving the desired result.


Previous answer: There exists $a\in (0,1)$ such that $f(a) = 1/2.$ Suppose $a\in (0,1/2].$ Then by Taylor,

$$ \frac{1}{2} = f(a) = f(0)+f'(0)\cdot a + f''(c)\frac{a^2}{2}=f''(c)\frac{a^2}{2}$$

for some $c\in (0,a).$ Thus $f''(c) = 1/a^2.$ Since $a^2\le 1/4,$ we see $f''(c)\ge 4.$ A similar proof works if $a\in [1/2,1).$

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Assume that $f''(x)< 4$, for all $x\le 1/2$. Then $$ f'(x)=f'(0)+\int_0^x f''(t)\,dt< 4x, $$ and $$ f(x)=f(0)+\int_0^x f'(t)\,dt< 2x^2\le\frac{1}{2} $$ which implies that $$ f\left(\frac{1}{2}\right)<\frac{1}{2} \tag{1} $$

If $f''(x)>-4$, for all $x\ge 1/2$, then $$ 0=f'(1)=f'(x)+\int_x^1 f''(t)\,dt> f'(x)-4(1-x), $$ and thus $$ -f'(x)>-4(1-x) $$ which implies that $$ f(x)-f(1)=-\int_x^1f'(t)\,dt>-4\int_x^1(1-t)\,dt=-2(1-x)^2, $$ and hence $$ f(x)>1-2(1-x)^2, $$ which implies that $$ f\left(\frac{1}{2}\right)>\frac{1}{2} \tag{2} $$ Observe that $(1)$ contradicts $(2)$.

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    $\begingroup$ We aren't assuming $f\in C^2[0,1],$ just that $f$ is twice differentiable on $[0,1].$ So it's possible $f''$ is not Riemann integrable. $\endgroup$ – zhw. Jun 21 '17 at 17:32
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    $\begingroup$ Your solution is very elegant, but I don't think that this is the solution that was envisioned as integrals do not appear in the text until 2 chapters after the problem is stated. Is there a way to do the problem using MVT or GMVT? $\endgroup$ – TheRealMathDaddy Jun 21 '17 at 17:32

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