Two inequalities for $\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}}$ 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!
 A: 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}
A: 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
