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Definitions: Let $\mathcal{P}_n$ be the vector space of all polynomials of degree $n$ or smaller with real-valued coefficients, thus $\mathcal{P}_n:= \left\{ \sum_{k=0}^n \lambda_k x^k\,\vert\,\lambda_k \in \mathbb{R}\right\}$ and a fixed polynom $\mathfrak{q}\in\mathcal{P}_n$ such as $\mathfrak{q}(x) = \sum_{k=0}^m\mu_kx^k$

Now we define a function $f\,:\,\mathcal{P}_n\longrightarrow\mathbb{R}$ with $f(\mathfrak{p}):= \mathfrak{q}(\mathfrak{p}(1))$

Question: I want to know if $\boldsymbol{f}$ is a differentiable function.

Attempt: I have already shown that the directional derivative exists for any arbitrary polynom $\mathfrak{r}\in\mathcal{P}_n$ and that it holds some kind of chain-rule like $\frac{\partial f}{\partial \mathfrak{r}}(\mathfrak{p}) = \mathfrak{q}'(\mathfrak{p}(1))\cdot\mathfrak{r}(1)$

Solution: I started on the left sight and I got $\frac{\partial f}{\partial \mathfrak{r}}(\mathfrak{p}) =\lim_{t\rightarrow 0}\frac{f(\mathfrak{p}+t\mathfrak{r})-f(\mathfrak{p})}{t} =\lim_{t\rightarrow 0} \frac{\sum_{k=0}^m\mu_k\left(\mathfrak{p}(1)+t\mathfrak{r}(1)\right)^k-\sum_{k=0}^m\mu_k\left(\mathfrak{p}(1)\right)^k}{t}$

Binomial expansion of the first sum should give something like: $\sum_{k=0}^m\mu_k\left(\mathfrak{p}(1)+t\mathfrak{r}(1)\right)^k = \sum_{k=0}^{m}\mu_k\left(\mathfrak{p}(1)^k+{k\choose 1}\mathfrak{p}(1)^{k-1}t\mathfrak{r}(1)+\mathcal{O}(t^2)\right)$

So finally I get: $\frac{\partial f}{\partial \mathfrak{r}}(\mathfrak{p}) = \sum_{k=1}^m k\mathfrak{p}(1)^{k-1}\mathfrak{r}(1) = \mathfrak{q}'(\mathfrak{p}(1))\cdot\mathfrak{r}(1)$

Question: Now all of those directional derivatives should be continous, aren't they? May I argue, that $f$ has to be differentiable then, even though i have not given a special basis for $\mathcal{P}_n$ or what else do I have to do?

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    Actually, for starters you might want to take your polynomials, say, on a segment $[0,1]$ and take any suitable norm for continous functions. However, a suggestion by @Berci looks a bit more suitable to your question.2012-12-03

0 Answers 0