Let $f(x)\in \mathcal C^2([a,b]),f(\frac{a+b}{2})=0$.

Show that$$\left|\int_{a}^{b}f(x)dx\right|\leq \frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx.$$

I try to use Taylor's Theorem with Integral form of the Remainder,then I have $$f(x)=f(\frac{a+b}{2})+f'(\frac{a+b}{2})(x-\frac{a+b}{2})+\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt.$$ Since $f(\frac{a+b}{2})=\int_{a}^{b}f'(\frac{a+b}{2})(x-\frac{a+b}{2})dx=0, $ we get$\int_{a}^{b}f(x)dx=\int_{a}^{b}\left(\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt\right)dx.$ I think the point of this problem is to prove: $$\left|\int_{a}^{b}\left(\int_{\frac{a+b}{2}}^{x}(x-t)f''(t)dt\right)dx\right|\leq\frac{1}{8}(b-a)^{2}\int_{a}^{b}\left|f''(x)\right|dx.$$The coeffient of $\frac{1}{8}$ is strange ,and how find it ?

  • $\begingroup$ how are you getting $$\int_a^b f'\left({a+b\over 2}\right)\left(x-{a+b\over 2}\right)\,dx=f\left({a+b\over 2}\right)\text{?}$$ $\endgroup$ – Adam Hughes Aug 6 '15 at 7:51
  • $\begingroup$ @AdamHughes The premise that $f\bigl(\frac{a+b}{2}\bigr) = 0$. $\endgroup$ – Daniel Fischer Aug 6 '15 at 7:55
  • $\begingroup$ @DanielFischer Oh I see, he's not appealing to properties of $f$ he's just using average value of a linear function. $\endgroup$ – Adam Hughes Aug 6 '15 at 7:56

Changing the order of integration, we obtain

\begin{align} \int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx &= \int_{\frac{a+b}{2}}^b \int_t^b (x-t)\,dx\, f''(t)\,dt\\ &= \int_{\frac{a+b}{2}}^b \frac{(b-t)^2}{2}f''(t)\,dt. \end{align}

Now we can take the absolute value to get

\begin{align} \biggl\lvert \int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert &\leqslant \int_{\frac{a+b}{2}}^b \frac{(b-t)^2}{2} \lvert f''(t)\rvert\,dt\\ &\leqslant \frac{(b-a)^2}{8} \int_{\frac{a+b}{2}}^b \lvert f''(t)\rvert\,dt, \end{align}

since $0 \leqslant b-t \leqslant b - \frac{a+b}{2} = \frac{b-a}{2}$ for $\frac{a+b}{2} \leqslant t \leqslant b$.

The analogous estimate for the integral over the first half of the interval gives the desired conclusion

\begin{align} \biggl\lvert \int_{a}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx\biggr\rvert &\leqslant \biggl\lvert \int_{\frac{a+b}{2}}^b \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert + \biggl\lvert \int_a^{\frac{a+b}{2}} \int_{\frac{a+b}{2}}^x (x-t)f''(t)\,dt\,dx \biggr\rvert\\ &\leqslant \frac{(b-a)^2}{8} \Biggl(\int_{\frac{a+b}{2}}^b \lvert f''(t)\rvert\,dt + \int_a^{\frac{a+b}{2}}\lvert f''(t)\rvert\,dt\Biggr). \end{align}

  • $\begingroup$ It is a pity for me I have not thought that $\int_{a}^{b}(\int_{a}^{x}f(x,y)dy)dx=\int_{a}^{b}(\int_{y}^{b}f(x,y)dx)dy.$ Your answer is very useful for me ! Thanks! $\endgroup$ – user202406 Aug 6 '15 at 9:13

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