Studying the differentiability of $f(P) = P(0)P'(1)$

Question

Let $E= \mathbb{R}_n[X]$ and $f: E \rightarrow \mathbb{R}$ such that $f(P) = P(0)P'(1)$. Show that it is differentiable at every $P \in E$

My teacher proposed a solution without giving full justification, which doesn't allow me to fully understand why it works. And also, I am unsure which norm to use.

He states that if we study for $P, H \in E$, we have:

$$f(P+H) = (P+H)(0)(P+H)'(1) = P(0)P'(1) + P(0)H'(1) + H(0)P'(1) + H(0)H'(1) $$

Thus we can put $df(P)(H) = P(0)H'(1) + H(0)P'(1)$ (which is linear) and as $H(0)H'(1)$, we get the differentiability.

Yet, if I study (with the condition that $H \not = 0$), we have: $$\frac{||f(P+H)-f(P) - df(P)(H)||}{||H||} = \frac{||H(0)H'(1)||}{||H||} $$ it doesn't seem that this would have a limit equal to $0$ when $||H|| \rightarrow 0$

Answer 1

Since $E$ is a finite-dimensional vector space, you can pick the norm you want, here a good choice is: $$\|P\|:=\sup_{x\in[0,1]}\max\{|P(x)|,|P'(x)|\},$$ you can check this is a well-defined norm, notice that only the separability axiom needs to be worked out. This way, one has: $$\frac{|H(0)H'(1)|}{\|H\|}\leqslant\|H\|,$$ which gives the result right away.

Answer 2

Whatever norm you pick, the continuous(!) maps $H\mapsto |H(0)|$ and $H\mapsto |H'(1)|$ are bounded on the compact(!) unit (under that norm) sphere, hence $|H(0)|\le c_1\|H\|$ and $|H'(1)|\le c_2\|H\|$ on that sphere and, by linearity, for all $H$. This makes $$\frac{|H(0)H'(1)|}{\|H\|}\le c_1c_2\|H\|\to 0.$$ Apparently, the same argument could be applied to any product of at least two linear functionals.