Let $T=[0,h]$ for some $h>0$. Show that, given $p\in[1,\infty]$ and $0\le m \le r$, there exists a constant $C$ such that for any $u\in C^\infty(T)$, there exists a polynomial $v\in P_r(T)$ satisfying $$|u-v|_{m,p,T}\le C|u|_{r+1,p,T}.$$
Find the exact value of the constant $C.$
My working has brought me to this $$|u-v|_{m,p,T}=\Bigl(\int^h_0\Bigl|\int^x_0{(x-t)^{r-m}\over (r-m)!}u^{(r+1)}(t)dt\Bigr|^{p}\Bigr)^{1\over p}$$
But I am stuck here. Your assistance is greatly appreciated.
Pull in absolute value, apply Hoelder in the inner integral, replace $\int_0^x$ by $\int_0^h$ to obtain $$ |u-v|_{m,p,T}^p=\int^h_0\Bigl|\int^x_0{(x-t)^{r-m}\over (r-m)!}u^{(r+1)}(t)dt\Bigr|^{p}dx\\ \le \int^h_0 \Bigl(\int^x_0{|x-t|^{r-m}\over (r-m)!}|u^{(r+1)}(t)|dt\Bigr)^{p}dx\\ \le \int^h_0 \Bigl(\int^x_0 \left({|x-t|^{r-m}\over (r-m)!}\right)^{\frac p{p-1}}dt\Bigr)^{p-1} \Bigl(\int^x_0 |u^{(r+1)}(t)|^p dt\Bigr) dx\\ \le \int^h_0 \Bigl(\int^x_0 \left({|x-t|^{r-m}\over (r-m)!}\right)^{\frac p{p-1}}dt\Bigr)^{p-1} dx \cdot |u|_{r+1,p,T}^p $$ It remains to evaluate the integral. Good luck! One idea would be to estimate $|x-t|\le h$, then this yields $$ C^p = \frac1{((r-m)!)^p} h^{\left((r-m)\frac p{p-1}+1\right)(p-1)+1}=\frac1{((r-m)!)^p} h^{(r-m+1)p} $$
