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Let $f$ be a continuous function on $[-1,1]$ and twice differentiable on $(-1,1)$.

Suppose that, $$\int_{-1}^{1} f(x) dx> f(1)+ f(-1)$$ Prove that there exists a number $x_0\in (-1,1)$ such that $f''(x_0)<0$.

I was told to consider $$\int_{-1}^{1} g(x) dx$$ where g is the function whose graph represents the line through $(-1 , f(-1))$ and $(1, f(1))$.

However, I'm not exactly sure how to utilise this information to the question above.

Thanks for help.

Community
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Buzzy123
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3 Answers3

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By Trapezoid rule error analysis, $$\exists \;c \in (a,b)\;:\; \int_{a}^{b}f(x)\,dx - \frac{b - a}{2}\cdot(f(a) + f(b))= - \frac{1}{12}f''(c)(b-a)^3.$$ Now let $a=-1$ and $b=1$. Can you take it from here?

P.S. Another reference: How prove this $\int_{a}^{b}f(x)dx=\frac{1}{2}(b-a)[f(a)+f(b)]-\frac{1}{12}(b-a)^3f''(\xi)$

Robert Z
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  • I'm sorry but I haven't learn this theorem in my calculus module. Is there another way to prove this? – Buzzy123 Oct 21 '19 at 01:54
  • @Buzz Well, I gave you a couple of links for the proof of that theorem. See https://math.stackexchange.com/q/535304 – Robert Z Oct 21 '19 at 06:18
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Hint: if you define the function $g$ as given then $$ \int_{-1}^1g(x)\,dx=f(1)+f(-1). $$ Now consider $h(x)=f(x)-g(x)$. You have $h(-1)=h(1)=0$ and $$ \int_{-1}^1 h(x)\,dx>0. $$ Can you say anything about existence of local maxima of $h$ on $[-1,1]$?

Hint (cont'd) : the mean value theorem for definite integrals ensures that there exists $c\in(-1,1)$ where $$ h(c)=\frac12\int_{-1}^1 h(x)\,dx>0. $$ Another application of the mean value theorem gives two points $a\in(-1,c)$ and $b\in(c,1)$ where $h'(a)>0$ and $h'(b)<0$. Can you say anything about $h''$ in $(a,b)$?

A.Γ.
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  • Thanks for your insight. Why is h(-1) = h(1) = 0? – Buzzy123 Oct 20 '19 at 12:13
  • @Buzz It follows from the definition of $g$ as the line that comes through the same points as the graph of $f$ at $-1$ and $1$. – A.Γ. Oct 20 '19 at 13:18
  • Alright I understand why but I'm not exactly sure about the significance of the area between f(x) - g(x) = h(x) > 0. Could you perhaps enlighten me? – Buzzy123 Oct 21 '19 at 09:52
  • @Buzz The mean value theorem several times could be one way to finish (see the edit). – A.Γ. Oct 21 '19 at 10:49
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Hint: suppose $f''(x) \geq 0$ for all $x$. Then $f$ is a convex function. This implies that $f \leq g$. It is also easy to check that $\int_{-1}^{1} g(x)dx=g(1)+g(-1)$. Hence $\int_{-1}^{1} g(x)dx=f(1)+f(-1)$. Thus $\int_{-1}^{1} f(x)dx \leq \int_{-1}^{1} g(x)dx=f(1)+f(-1)$. This contradicts the hypothesis.

Proof of the fact that $f \leq g$: let $x \in [-1,1]%=$. Then we can write $x=(t)(1)+(1-t)(-1)$ for some $t \in [0,1]$. Hence $f(x) \leq t f(1)+(1-t)f(-1)=t g(1)+(1-t)g(-1)=g(x)$ since $g$ is an affine function.

Kavi Rama Murthy
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