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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.

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    $\begingroup$ Why must the cases $f''(x)\equiv1$ and $f''(x)=-1$ necessarily give the extreme values for $\int_0^1|f(x)|\,dx$? $\endgroup$ Aug 2, 2020 at 8:27
  • $\begingroup$ @AnginaSeng Not sure. It just felt like the natural thing to assume. It actually works out if you assume $f''(x)=x$. $\endgroup$
    – C Squared
    Aug 2, 2020 at 8:38
  • $\begingroup$ "Not sure." This is why I only see here a "plausibility argument" but not a proof. $\endgroup$ Aug 2, 2020 at 8:41
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    $\begingroup$ 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 $f’’= \pm 1$. $\endgroup$
    – Aphelli
    Aug 2, 2020 at 8:41
  • $\begingroup$ @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. $\endgroup$
    – C Squared
    Aug 2, 2020 at 8:43

2 Answers 2

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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.

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

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  • $\begingroup$ @Csquared is there a problem so that you unaccepted the answer? $\endgroup$
    – VIVID
    Aug 2, 2020 at 9:12
  • $\begingroup$ 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. $\endgroup$
    – C Squared
    Aug 2, 2020 at 9:13
  • $\begingroup$ @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... $\endgroup$
    – VIVID
    Aug 2, 2020 at 9:15
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    $\begingroup$ @CSquared VIVID's solution gives an upper bound on the absolute value of the integral. That's all the problem asks to be shown. $\endgroup$
    – user635640
    Aug 2, 2020 at 9:21
  • $\begingroup$ @CSquared VIVID has shown $|\int f|\le\int|g|=\frac{1}{12}$, and $g''=\pm1$. $\endgroup$ Aug 2, 2020 at 9:23
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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.$$\left| \int_0^1{f\left( x \right) \text{d}x} \right|\leqslant \frac{1}{12}$$

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