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Assume $f$ has a continuous second derivative with $f(0) = f'(0) = f'(1) = 0$ and $f(1) = 1$. Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$.

We must also have that $f'$ is continuous by differentiability. Therefore, by Rolles theorem there exists a $c$ in $[0,1]$ such that $f''(c) = 0$. I am not sure how to use the fact that $f(1) = 1$ to show that $|f''(x)| > 4$.

Puzzled417
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  • How about using Taylor series expansion? – Larara Mar 23 '16 at 22:48
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    @Larara What if $f$ doesn't have a convergent Taylor series expansion? – Puzzled417 Mar 23 '16 at 22:53
  • The average slope of $f(x)$ over the interval must be $1$, and therefore there must be a $c_1$ such that $f'(c_1)\ge 1$; you can probably prove that such a point must occur at a point $c_2$ where $f''(c_2)=0$, and you can again apply logic, this time to say that the average slope of $f'(x)$ must be greater than $4$ since the interval is at most $\frac14$ of the length of the original... – abiessu Mar 23 '16 at 23:03

3 Answers3

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$$f(\frac{1}{2}) = f(0) + f'(0)(\frac{1}{2}) + \frac{f''(\theta_1)}{2}(\frac{1}{2})^2 = f''(\theta_1)\frac{1}{8}$$ $$f(\frac{1}{2}) = f(1) - f'(1)(\frac{1}{2}) + \frac{f''(\theta_2)}{2}(\frac{1}{2})^2 = 1 + f''(\theta_2)\frac{1}{8}$$ Here, $0\le \theta_1 \le 1/2$, and $1/2 \le \theta_2 \le 1$ Combining both, we have $$ 1 = f''(\theta_1)\frac{1}{8} - f''(\theta_2)\frac{1}{8}$$ Take absolute value, $$ 1 \le max \{|f''(\theta_1)|, |f''(\theta_2)|\} \frac{1}{4}$$

So c is one of $\theta_1, \theta_2$.

Let $$f(x) = \begin{cases} 2x^2\, \text{ for } 0 \le x \le 1/2\\ -2(x-1)^2 +1\, \text{ for } 1/2 < x \le 1 \end{cases}$$

$$f''(x) = 4$$, so the equality holds for this f.

runaround
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We'll prove strict inequality by contradiction.

Set $g=f'$. By hypothesis, $g$ is $C^1$ on $[0,1]$, $\;g(0)=g(1)=0$ and $\displaystyle \int_0^1g(x)\,\mathrm d\mkern1mu x=1$.

Suppose $\lvert g'(x)\rvert\le 4$. This means

  • $g'\le 4$, which implies $g(x)\le 4x$ on $\bigl[0,\frac12\bigr]$ so that $$\int_0^{1/2}\!\!g(x)\,\mathrm d\mkern1mu x\le4\int_0^{1/2}x\,\mathrm d\mkern1mu x=\frac12.$$ Furthermore, we have equality if and only if $g(x)=4x$ on $\bigl[0,\frac12\bigr]$.
  • $g'\ge -4$, which implies $g(x)\le 4(1-x)$ on $\bigl[\frac12,1\bigr]$ so that $$\int_{1/2}^1g(x)\,\mathrm d\mkern1mu x\le4\int_{1/2}^1(1-x)\,\mathrm d\mkern1mu x=\frac12,$$ and we have equality if and only if $ g(x)=4((1-x) $ on $\bigl[\frac12,1\bigr]$.

Now, as $\displaystyle \int_0^1g(x)\,\mathrm d\mkern1mu x=1$, we do have equality in both cases. Thus $g$ is defined by $$g(x)=\begin{cases}4x &\text{if}\enspace x\in\bigl[0,\frac12\bigr],\\ 4(1-x)&\text{if}\enspace x\in\bigl[\frac12,1\bigr]. \end{cases}$$ This contradicts derivability of $g$ at $x=\frac12$.

Bernard
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  • Not at all: the problem stipulates $f$ is $C^2$, hence $f'$ is $C^1$. – Bernard Mar 24 '16 at 00:09
  • Yes because at $x=\frac12$, $g$ has a left derivative equal to $4$ and a right derivative equal to $-4$. This contradicts being differentiable on $[0,1]$. – Bernard Mar 24 '16 at 00:16
  • But that might mean that $g' < -4$? Also I don't see how taking the equality case in both cases can make a contradiction. – Puzzled417 Mar 24 '16 at 00:18
  • The curve is not smooth at $x=\frac12$ since it has only (different) one sided derivatives. The hypotheses say it does have a (two-sided) derivative. That is contradiction: it both has and has not a derivative at a point. – Bernard Mar 24 '16 at 00:22
  • How do you know in the first case equality is achieved if and only if $g(x) = 4x$? – Puzzled417 Mar 24 '16 at 00:30
  • This means $\int_0^{1/2}\bigl(g(x)-4x\bigr),\mathrm d\mkern1mu x=0$, and $g(x)-4x$ is continuous, $\ge 0$. This implies $g(x)-4x=0$. – Bernard Mar 24 '16 at 00:32
  • I think you mean that since $g(x) - 4x \leq 0$, we must have $g(x) = 4x.$ – Puzzled417 Mar 24 '16 at 00:42
  • Oh! Yes, sorry for the typo. – Bernard Mar 24 '16 at 00:52
  • One more thing: how do you know $g(x) \leq 4x$ on $[0,1/2]$ and not $[0,1]$? – Puzzled417 Mar 24 '16 at 00:58
  • I didn't say it was not true on $[0,1]$. I use the inequality on the interval which is useful in the context. – Bernard Mar 24 '16 at 00:59
  • Nice solution! This does assume, though, that such a function exists right? Otherwise our proof may be invalid. – Puzzled417 Mar 24 '16 at 01:08
  • You can easily sketch one by hand, hence it must exist (take a suitable Bézier curve). It's a classical, not-so-easy exercise. – Bernard Mar 24 '16 at 01:11
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We have $$ \eqalign{ 1&=f(1)-f(0)=\int_0^1 f'(x)\,dx\cr &=\int_0^{1/2}f'(x)\,dx+\int_{1/2}^1 f'(x)\,dx\cr &=\int_0^{1/2}\int_0^xf''(t)\,dt\,dx-\int_{1/2}^1 \int_x^1f''(t)\,dt\,dx\cr &=\int_0^{1/2}(1/2-t)f''(t)\,dt-\int_{1/2}^1 (t-1/2)f''(t)\,dt.\cr } $$ Consequently, if $|f''(t)|\le 4$ for all $t\in[0,1]$, then by the above and the triangle inequality, $$ \eqalign{ 1&\le 4 \int_0^{1/2}(1/2-t)\,dt+4\int_{1/2}^1(t-1/2)\,dt\cr &=-2(1/2-t)^2\Big|_0^{1/2}+2(t-1/2)^2\Big|_{1/2}^1\cr &=1/2+1/2=1.\cr } $$ It follows that $$ \int_0^{1/2}(1/2-t)|f''(t)|\,dt=\int_{1/2}^1 (t-1/2)|f''(t)|\,dt=1/2, $$ which togetther with the hypothesis $|f''(t)|\le 4$ for all $t$ implies that $|f''(t)|=4$ for all $t\in[0,1]$. As $f''$ is continuous, we must have either $f''(t)=4$ for all $t$ or $f''(t)=-4$ for all $t$. In the first case $f(x)=2x^2$ in violation of $f(1)=1$. Likewise the second alternative leads to a contradiction.

John Dawkins
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