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Suppose that $f\in C^2$, $f''(x)\geq 0$ $\,\,\,\forall x \in [a,b]$. I want to show that $$\frac{1}{2}(b-a)(f(a)+f(b))\leq \int_a^bf(t)\,dt\leq (b-a)f\left(\frac{a+b}{2}\right).$$If we divide by $b-a$, we see that the left term is less than the right term by definition of convexity, and it remains to show that the average value of the function lies between $f\left(\frac{a+b}{2}\right)$ and $\frac{1}{2}(f(a)+f(b))$.

The mean value theorem for integrals implies that the average value is attained at some point $c\in (a,b)$. But it's not clear to me why $f(c)$ should lie in between two other points. Perhaps there's another theorem about integration we should apply. Any ideas?

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I think the inequality needs to be reversed.

$\displaystyle \frac{b-a}{2}f\Big(\frac{b+a}{2}\Big)-\int_a^{\frac{b+a}{2}}f(t)\,dt=\int_a^{\frac{b+a}{2}}\int_t^{\frac{b+a}{2}}f'(s)\,ds\,dt\le\int_{\frac{b+a}{2}}^{b}\int_{\frac{b+a}{2}}^{t}f'(s)\,ds\,dt$ (since $f'(s)$ is bigger in the latter region and the integration is over a domain of the same size) $\displaystyle=\int_{\frac{b+a}{2}}^{b}f(t)\,dt-\frac{b-a}{2}f\Big(\frac{b+a}{2}\Big),$ hence $\displaystyle (b-a)f(\frac{b+a}{2})\le\int_a^b f(t)\,dt$.

Other side:

$\displaystyle \int_a^{\frac{b+a}{2}}f(t)\,dt-\frac{b-a}{2}f(a)=\int_a^{\frac{b+a}{2}}\int_a^{t}f'(s)\,ds\,dt\le \int_{\frac{b+a}{2}}^{b}\int_{t}^{b}f'(s)\,ds\,dt=\frac{b-a}{2}f(b)-\int_{\frac{b+a}{2}}^b f(t)\,dt\,,$ hence $\displaystyle \int_a^b f(t)\,dt\le\frac{b-a}{2}(f(b)+f(a))\,.$

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  • $\begingroup$ You're right; I meant to write $f''\leq 0$. But I can just reverse your argument! $\endgroup$ – Eric Auld Jul 9 '13 at 3:36
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    $\begingroup$ For the first inequality in the "other side" part, the inner integral on the left won't be $<=$ the one on the right if $t = b$ for example, so you maybe are comparing the inner integral at $t$ with the integral at $a + b - t$? Maybe you should clarify this... $\endgroup$ – Zarrax Jul 9 '13 at 14:48
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    $\begingroup$ The $t$ on the left runs from $a$ to $\frac{b+a}{2}$ and the one on the right runs from $\frac{b+a}{2}$ to $b$. The left integral is bounded above by $\frac{(b-a)^2}{8}\sup\limits_{t\in\left(a,\frac{b+a}{2}\right)}f'(t)$ which is clearly less than the integral on the right. $\endgroup$ – JLA Jul 9 '13 at 17:23
  • $\begingroup$ (woops edit time ran out) The thing in the box should be $\frac{(b-a)^2}{8}\sup_{t\in(a,\frac{b+a}{2})}f'(t)$, and this is a lower bound for the integral on the right. $\endgroup$ – JLA Jul 9 '13 at 17:29
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Here is an answer that doesn't assume $f$ is differentiable.

$f$ is convex iff $\frac{f(x)-f(y)}{x-y}$ is non-decreasing in both $x$ and $y$. If $f$ is convex, $f$ is continuous.

Assume $a\lt b$ and let $c=\frac{a+b}{2}$.

Since $\dfrac{f(x)-f(c)}{x-c}$ is non-decreasing, $$ D^-=\sup_{x\lt c}\frac{f(x)-f(c)}{x-c} \le\inf_{x\gt c}\frac{f(x)-f(c)}{x-c}=D^+\tag{1} $$ Let $D=\frac{D^-{+}D^+}{2}$.

Inequality $(1)$ says $$ f(x)-f(c)\ge D(x-c)\tag{2} $$ Integrate $(2)$ over $[a,b]$ to get $$ \begin{align} \int_a^bf(x)\,\mathrm{d}x-(b-a)f(c) &\ge D\left[\frac12x^2-cx\right]_a^b\\ &=D\left[\frac12(b^2-a^2)-\frac{a+b}2(b-a)\right]\\[6pt] &=0\tag{3} \end{align} $$ For any $x\in[a,b]$, the convexity of $f$ implies $$ \frac{f(x)-f(a)}{x-a}\le\frac{f(x)-f(b)}{x-b}\tag{4} $$ therefore, since $(x-a)(x-b)\le0$, $(4)$ is equivalent to $$ (x-b)(f(x)-f(a))\ge(x-a)(f(x)-f(b))\tag{5} $$ Subtract $xf(x)$ from both sides of $(5)$ and integrate to get $$ \frac12(b^2-a^2)(f(b)-f(a))+(bf(a)-af(b))(b-a) \ge(b-a)\int_a^bf(x)\,\mathrm{d}x\tag{6} $$ Divide both sides of $(6)$ by $b-a$ and regroup to get $$ \begin{align} \int_a^bf(x)\,\mathrm{d}x &\le\frac12(a+b)(f(b)-f(a))+(bf(a)-af(b))\\ &=\frac12(b-a)(f(a)+f(b))\tag{7} \end{align} $$ Bringing $(3)$ and $(7)$ together yields $$ (b-a)f\left(\frac{a+b}{2}\right) \le\int_a^bf(x)\,\mathrm{d}x \le\frac12(b-a)(f(a)+f(b))\tag{8} $$

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By a change of variable $t \rightarrow a + b - t$, you have $\int_a^bf(t)\,dt = \int_a^bf(a + b - t)\,dt$, so what you're trying to prove is equivalent to $$(b-a)(f(a)+f(b))\leq \int_a^b(f(t) + f(a + b - t))\,dt\leq 2(b-a)f\left(\frac{a+b}{2}\right)$$ It then suffices to prove that for each $a < t < b$ that $$f(a)+f(b)\leq f(t) + f(a + b - t)\leq 2f\left(\frac{a+b}{2}\right)$$ The result will then follow by integrating in $t$ from $a$ to $b$. By symmetry it suffices to just look at $a < t < {a + b \over 2}$. The left inequality can be rewritten as $$f(b) - f(a + b - t) \leq f(t) - f(a)$$ By the mean-value theorem, the left-hand side is $(t - a)f'(c)$ for some $c > a + b - t > {a + b \over 2}$, and the right-hand side is $(t-a)f'(d)$ for some $d < t < {a + b \over 2}$. Thus the fact that $f'' \leq 0$ gives the inequality.

Similarly, the right-hand inequality can be rewritten as $$f(a + b - t) - f(\frac{a + b}{2}) \leq f(\frac{a + b}{2}) - f(t)$$ The left-hand side is $(\frac{a + b}{2} - t)f'(d)$ for some $d > {a + b \over 2}$, and the right-hand side is $(\frac{a + b}{2} - t)f'(c)$ with $c < {a + b \over 2}$, so the condition that $f'' \leq 0$ gives the right-hand inequality as well.

Note you only need that $f'$ is decreasing, you don't actually need the second derivative to exist.

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