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Given that a function $f$ has a continuous second derivative on the interval $[0,1]$, $f(0)=f(1)=0$, and $|f''(x)|\leq 1$, show that $$\left|\int_{0}^{1}f(x)\,dx\right|\leq \frac{1}{12}\,.$$

My attempt: This looks to be a maximization/minimization problem. Since the largest value $f''(x)$ can take on is $1$, then the first case will be to assume $f''(x)=1$. This is because it is the maximum concavity and covers the most amount of area from $[0,1]$ while still maintaining the given conditions.

Edit: Because of the MVT and Rolle's Theorem, there exists extrema on the interval $[0,1]$ satisfying $f'(c)=0$ for some $c\in[0,1]$. These extrema could occur at endpoints.

Then $f'(x)=x+b$ and $f(x)=\frac{x^2}{2}+bx+c$. Since $f(0)=0$, then $c=0$ and $f(1)=0$, then $b=-\frac{1}{2}$. Remark: Any function with a continuous, constant second derivative will be of the form $ax^2+bx+c$ and in this case, $a=-b$ and $c=0$. Now, $$\begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(x^2-x)\,dx\\&=\frac{1}{2}\bigg[\frac{x^3}{3}-\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=-\frac{1}{12}\end{align*}$$

Next, we assume that $f''(x)=-1$ and repeating the process yields $$ \begin{align*}\int_{0}^{1}f(x)\,dx&=\frac{1}{2}\int_{0}^{1}(-x^2+x)\,dx\\&=\frac{1}{2}\bigg[\frac{-x^3}{3}+\frac{x^2}{2}\bigg]_{x=0}^{x=1}\\&=\frac{1}{12}\end{align*}$$ Thus we have shown that at the upper and lower bounds for $f''(x)$ that $\frac{-1}{12}\leq\int_{0}^{1}f(x)\,dx\leq \frac{1}{12} \Longleftrightarrow \left|\int_{0}^{1}f(x)\,dx\right|\leq\frac{1}{12}$ because $f''(x)$ is continuous on $[0,1]$.

I was wondering if this was 'rigorous' enough to be considered a full proof and solution to the problem.

VIVID
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C Squared
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    Why must the cases $f''(x)\equiv1$ and $f''(x)=-1$ necessarily give the extreme values for $\int_0^1|f(x)|,dx$? – Angina Seng Aug 02 '20 at 08:27
  • @AnginaSeng Not sure. It just felt like the natural thing to assume. It actually works out if you assume $f''(x)=x$. – C Squared Aug 02 '20 at 08:38
  • "Not sure." This is why I only see here a "plausibility argument" but not a proof. – Angina Seng Aug 02 '20 at 08:41
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    I’m sorry, but it’s not. You don’t prove the existence of a maximum (which isn’t obvious), you don’t provide a satisfactory explanation about why said maximum must satisfy $f’’= \pm 1$. – Aphelli Aug 02 '20 at 08:41
  • @AnginaSeng Yea. I think if $f''(x)$ is any function which meets the criteria, then the inequality will hold. You just have to find what input value gives the maximum and minimum output values for the second derivative on the interval. So my choice of a constant function was in the set of valid functions. – C Squared Aug 02 '20 at 08:43
  • @Mindlack how would one go about doing that? By just considering a function that exists and meets the conditions? – C Squared Aug 02 '20 at 08:46
  • @Mindlack C Squared has the right intuition. Though it is not a proof, it is correct to try extreme values by Pontryagin's maximum principle. – Chrystomath Aug 02 '20 at 08:49
  • @Crystomath: I’m not denying that it’s a good idea to try extreme values. I’m just saying that, a priori, there’s no given rigorous argument why 1) they would be optimal; 2) they would be the only optimal solutions; 3) infinite-dimension phenomena (eg it’s a sup, not a max) cannot occur. All three issues can probably be resolved, of course. – Aphelli Aug 02 '20 at 09:17

2 Answers2

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Consider the following integral: $$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx. $$

By integrating by parts twice, you get

$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx = \underbrace{\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f'(x)\bigg|_0^1}_{0} - \int_0^1\bigg(x-\frac{1}{2}\bigg)f'(x)dx=$$$$= - \int_0^1\bigg(x-\frac{1}{2}\bigg)f'(x)dx= \underbrace{- \bigg(x-\frac{1}{2}\bigg)f(x)\bigg|_0^1}_{0} + \int_0^1f(x)dx$$ Therefore, $$\boxed{\int_{0}^{1}f(x)\, dx = \int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx}$$

Now use the following inequality: $$\left|\int_{a}^{b}f(x)g(x)\,dx\right| \leq \int_{a}^{b}|f(x)||g(x)|\, dx$$

Since $g(x)=\frac{x^{2}}{2}-\frac{x}{2}$ is the expression you got, this should yield the desired result.

$$\left|\int_0^1 f(x)\,dx\right|=\left| \int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx\right|\le\frac{1}{2}\int_{0}^{1}|x^2-x|\,dx=\frac{1}{12}$$

VIVID
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  • @Csquared is there a problem so that you unaccepted the answer? – VIVID Aug 02 '20 at 09:12
  • Well, the problem is proving existence of a maximum and why the maximum must satisfy $f''(x)=\pm1$ You used part of my attempt, which uses an invalid claim. – C Squared Aug 02 '20 at 09:13
  • @CSquared You asked, "I was wondering if this was 'rigorous' enough to be considered a full proof and solution to the problem." and I think you got enough reactions for this in the comments above. And I wanted to provide a rigorous proof... – VIVID Aug 02 '20 at 09:15
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    @CSquared VIVID's solution gives an upper bound on the absolute value of the integral. That's all the problem asks to be shown. –  Aug 02 '20 at 09:21
  • @CSquared VIVID has shown $|\int f|\le\int|g|=\frac{1}{12}$, and $g''=\pm1$. – Chrystomath Aug 02 '20 at 09:23
  • @VIVID Super sorry, I didn't make the connection for the absolute value of $f''(x)=1$ in the last step. I guess I am still a bit confused on what you integrated by parts in the first step. I know you're not wrong, its just I can't get my answer to match up with yours. Did you differentiate $f(x)$ and integrate $1$? – C Squared Aug 02 '20 at 09:32
  • @CSquared Better try to prove that the RHS is equal to LHS by integrating by parts. – VIVID Aug 02 '20 at 09:54
  • @VIVID When I integrated by parts, the left hand side was $xf(x)-(f’(x)x^2)/2 +\int( f’’(x)x^2)/2 dx$. I am confused because I got $g(x)$ on the baseless assumption that $ f’’(x)=1$ yielded an extrema. – C Squared Aug 02 '20 at 10:09
  • @CSquared See the edit for the integrating process. – VIVID Aug 02 '20 at 10:46
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Use $\text{Taylor}$ we can get: $$ f\left( 0 \right) =f\left( x \right) -xf'\left( x \right) +\frac{f''\left( \xi _1 \right)}{2}x^2,\xi _1\in \left( 0,x \right) $$ $$ f\left( 1 \right) =f\left( x \right) +\left( 1-x \right) f'\left( x \right) +\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2,\xi _2\in \left( x,1 \right) $$ so we can get: $$0=2f\left( x \right) +\left( 1-2x \right) f'\left( x \right) +\frac{f''\left( \xi _1 \right)}{2}x^2+\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2$$ i.e.$$2f\left( x \right) =\left( 2x-1 \right) f'\left( x \right) -\frac{f''\left( \xi _1 \right)}{2}x^2-\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2$$ then we can get: $$2\int_0^1{f\left( x \right) \text{d}x}=\int_0^1{\left[ \left( 2x-1 \right) f'\left( x \right) -\frac{f''\left( \xi _1 \right)}{2}x^2-\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2 \right] \text{d}x}$$ after simplification,we can get:$$4\int_0^1{f\left( x \right) \text{d}x}=-\int_0^1{\left[ \frac{f''\left( \xi _1 \right)}{2}x^2+\frac{f''\left( \xi _2 \right)}{2}\left( x-1 \right) ^2 \right] \text{d}x}$$ so$$4\left| \int_0^1{f\left( x \right) \text{d}x} \right|\leqslant \frac{1}{2}\int_0^1{\left( 2x^2-2x+1 \right) \text{d}x}=\frac{1}{3}$$ i.e.$$\left| \int_0^1{f\left( x \right) \text{d}x} \right|\leqslant \frac{1}{12}$$

xuke0721
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