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Let $f$ be a $C^2$ function on $[0,1]$. $f(0)=f(1)=f'(0)=0,f'(1)=1.$ Prove that

$$\int_0^1|f''(x)| \, dx\ge4.$$

Also determine all possible $f$ when equality occurs.

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    $\begingroup$ Have you tried integrating by parts? $\endgroup$
    – user38268
    Oct 23, 2012 at 2:52

1 Answer 1

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The inequality does not hold. Take $$ f(x) = x^3 - x^2 $$ we have $$ f'(x) = 3x^2 - 2x \\ f''(x) = 6x - 2 \\ f(0) = f(1) = f'(0) = 0 \\ f'(1) = 1 $$ But $$ \int_0^1 \lvert f''(x)\rvert dx = \int_0^{1/3} (2 - 6x) dx + \int_{1/3}^1 (6x - 2)dx = 1/3 + 4/3 = 5/3 < 4 $$

Then, what can we say about the infimum of that integral? $$ \int_0^1 \lvert f''(x) \rvert dx \geq \int_0^1 f''(x) dx = f'(1) - f'(0) = 1 $$ To show that $1$ is the best lower bound, choose a $C^2$ function $h$ defined on $[0, 1/2]$ such that $$ h(0) = h'(0) = h'(1/2) = h''(1/2) = 0 \\ h(1/2) = -1 $$ Using the above function we can construct the map $$ f(x) = \begin{cases} kh(x) & \text{if }0\leq x \leq 1/2 \\ -k & \text{if }1/2 < x \leq 1 - 3k \\ \frac {(x - 1 +3k)^3}{27k^2} - k & \text{if }1 - 3k < x \leq 1 \end{cases} $$ where $k$ is a positive constant lesser than $1/6$. $f$ satisfies all the constraints and $$ \int_0^1 \lvert f''(x) \rvert dx = 1 + k\int_0^{1/2} \lvert h''(x) \rvert dx $$ Choosing $k$ small enough we can make the integral as close to $1$ as we want.

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  • $\begingroup$ Wow... how did you come up with this counter-example? $\endgroup$
    – wj32
    Oct 23, 2012 at 10:47
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    $\begingroup$ @wj32: Dear wj32, If you look for a function satisfying the conditions, this is the first one that comes to mind. Regards, $\endgroup$
    – Matt E
    Oct 23, 2012 at 11:02
  • $\begingroup$ Ok, but does the integral have an infimum? $\endgroup$ Oct 23, 2012 at 13:47
  • $\begingroup$ I added details about the infimum. $\endgroup$
    – AlbertH
    Oct 23, 2012 at 14:59

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