Let $n\geq 2$ an integer and $E=\{f\in\mathscr{C}^n([0,1],\mathbb{R}),\; f(0)=f\left(\frac 1{n-1}\right)=f\left(\frac 2{n-1}\right)=...=f(1)=0\}$, equipped with the norm $\|f\|=\displaystyle\int_0^1|f^{(n)}(t)|\text{d}t$. My aim is to find the best constant $C$ such that $\forall f\in E,\; \|f\|_\infty\leq C\|f\|$ and to know if the equality is possible for one function $f\neq 0$. My attempts, only for $n=2$ : if $f\in E$, there exists $c\in]0,1[$ where $|f|$ is at its largest, with $f'(c)=0$. Then $$\displaystyle f(0)=f(c)+f'(c)(0-c)+\int_c^0(0-t)f''(t)\text{d}t$$ and $$\displaystyle f(1)=f(c)+f'(c)(1-c)+\int_c^1(1-t)f''(t)\text{d}t,$$
thus, adding : $$2f(c)=\int_c^0tf''(t)\text{d}t+\int_c^1(t-1)f''(t)\text{d}t$$
which implies $\displaystyle\|f\|_\infty\leq\frac 12\|f\|$.
I feel that this constant $\frac 12$ is not the best (I expect $\frac 14$, but don't know how to take advantage of the terms $t$ and $(t-1)$ in the integrals). So :
- is $\frac 12$ the best constant for $n=2$ ? If not, what is the best constant, and is there equality for one $f\neq 0$ in $E$ ?
- what about $n\geq 3$ with the same questions ?
a sketch for $n=2$:
Using your approach, we obtain $$ f(c) = \int_0^c (-t)f''(t) \,\mathrm dt = \int_c^1 (1-t) f''(t)\,\mathrm dt. $$ This implies the estimate $$ \|f\|_\infty = |f(c) | \leq \min\left( c \int_0^c|f''(t)|\,\mathrm dt , (1-c) \int_c^1|f''(t)|\,\mathrm dt \right). $$ This estimate has to be strict if $f''$ is not constantly zero on both $(0,c)$ and $(c,1)$ (which has to be the case if $f(c)\neq0$).
Without loss of generality we assume that $\|f\|=1$ and we set $b=\int_0^c|f''(t)|\,\mathrm dt \in [0,1]$. Then we have $\|f\|_\infty < \min(cb,(1-c)(1-b))$, which can be shown to be bounded by $\tfrac14$.
However, one can come arbitrarily close to the constant $\tfrac14$ by using functions with $f''(x)= k/2$ on $(\tfrac12 - \tfrac1k,\tfrac12+\tfrac1k)$, and $f''(x)=0$ elsewhere (and constructing $f$ such that $f(1)=f(0)=0$).
Thus, $\tfrac14$ has to be the best constant.