Show that if $f\in \mathcal C^{n+1}([a,b])$ and $f(a)=f^{'}(a)=\cdots=f^\left(n\right)(a)=0,$ then the following statements are ture:

$\mathbf a)$

$ \forall r\in[1,\infty),$the inequality $$\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}} \leq \frac{(b-a)^{n+\frac{1}{r}}}{n!(nr+1)^{\frac{1}{r}}}\int_{a}^{b}|f^{(n+1)}(x)|dx$$holds.

$\mathbf b)$

$ \forall r\in[1,\infty),$the inequality $$\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}} \leq \frac{2^{\frac{1}{r}}(b-a)^{n+\frac{1}{r}+\frac{1}{2}}}{n!\sqrt{2n+1}(2nr+r+1)^{\frac{1}{r}}}\left(\int_{a}^{b}|f^{(n+1)}(x)|^{2}dx\right)^{\frac{1}{2}}$$holds.

Using Taylor's Theorem with Integral form of the Remainder,I can easily get $$\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}} \leq \frac{(b-a)^{n+\frac{1}{r}}}{n!}\int_{a}^{b}|f^{(n+1)}(x)|dx.\quad(\star)$$ I tried to apply Holder's inequality for integrals to $(\star) $ in question $\mathbf a)$,but I have yet to prove it holds.I believe this two questions might have the same method.Any help you can provide will be greatly appreciated!

  • $\begingroup$ Did you try to apply Taylor's Theorem straight to the function $f^r(x)$, which has also zero derivatives at $a$ up to order $n$ ? $\endgroup$ – Svetoslav Aug 6 '15 at 12:18
  • $\begingroup$ @Svetoslav ;oh,I think that it less effective. $\endgroup$ – user202406 Aug 6 '15 at 13:37
  • $\begingroup$ I think you are right. The $n+1$ derivative is tough to compute. $\endgroup$ – Svetoslav Aug 6 '15 at 16:34

Firstly, let's remark one application of these inequalities: they bound the different(in terms of $L^r([a,b])$ norm) between a smooth function and its Taylor approximation, since for any $g\in \mathcal C^{n+1}([a,b])$ and its Taylor approximation $p_n$ of order $n$, $g-p_n$ satisfies the given condition. Then we will prove these inequalities:

We have by Taylor's Theorem with Integral form of the Remainder \begin{align} f(x) = \int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt \end{align}

Then we have \begin{align} \int_a^b |f(x)|^rdx &= \int_a^b \left|\int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\right|^rdx \\ &\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right| \left|(x-t)^n \right|dt\right)^rdx \\ &\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt (x-a)^n\right)^rdx \\ &\leq \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt\right)^r\int_a^b \left( (x-a)^n\right)^rdx \\ & = \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt\right)^r\frac{(b-a)^{nr+1}}{nr+1} \\ \end{align}

So we get \begin{align} \left(\int_a^b |f(x)|^rdx\right)^{1/r} \leq \left(\frac{(b-a)^{nr+1}}{nr+1}\right)^{1/r} \int_a^b \left|\dfrac{f^{(n+1)}(x)}{n!}\right|dx \end{align} which is $\mathbf a)$

To get $\mathbf b)$ we can proceed similarly using Holder's inequality

\begin{align} \int_a^b |f(x)|^rdx &= \int_a^b \left|\int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\right|^rdx \\ &\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt \int_a^x \left|(x-t)^{2n} \right|dt\right)^{r/2} dx \\ &\leq \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \int_a^b \left(\int_a^x \left|(x-t)^{2n} \right|dt\right)^{r/2} dx \\ &= \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \int_a^b \left(\frac{(x-a)^{2n+1}}{2n+1}\right)^{r/2} dx\\ & = \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \left(\frac{1}{2n+1}\right)^{r/2} \left(\frac{(b-a)^{nr+\frac{r}{2} +1}}{nr+\frac{r}{2} +1}\right)\\ \end{align}

  • $\begingroup$ It is a nice answer made consistent with my idea,but I failed to go further.Thanks . $\endgroup$ – user202406 Aug 6 '15 at 13:45

There is an integration by part to get the $nr+1$ term:

$$\int_a^b |f(x)|^r dx = \int_a^b \left| \frac{(x-a)^n}{n!} \int_a^x f^{(n-1)}(t) dt \right|^r dx$$

$$\leq \int_a^b \frac{(x-a)^{nr}}{n!^r} \left| \int_a^x f^{(n-1)}(t) dt \right|^r dx $$

$$\leq \int_a^b \frac{(x-a)^{nr}}{n!^r} \left( \int_a^x \left| f^{(n-1)}(t) \right| dt \right)^r dx $$

$$\leq \left[ \frac{(x-a)^{nr+1}}{n!^r(nr+1)} \left( \int_a^x \left| f^{(n-1)}(t) \right| dt \right)^r \right]_a^b - \int_a^b \frac{(x-a)^{nr+1}}{n!^r(nr+1)} r \left| f^{(n-1)}(x) \right| \left( \int_a^x \left| f^{(n-1)}(t) \right| dt \right)^{r-1} dx $$

And as the second part is negative, you get

$$\leq \left[ \frac{(x-a)^{nr+1}}{n!^r(nr+1)} \left( \int_a^x \left| f^{(n-1)}(t) \right| dt \right)^r \right]_a^b =\frac{(b-a)^{nr+1}}{n!^r(nr+1)} \left( \int_a^b \left| f^{(n-1)}(t) \right| dt \right)^r $$

Hence the result


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