Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Why do you go for directional (Gateaux?) and not Frechet derivatives? – bonext Dec 3 '12 at 16:12
The idea was to show, that the partial derivatives are continous, so $f$ is a differentiable function. I calculated all directional derivatives because I do not know the basis of $\mathcal{P}_n$. I think i cannot use Frechet derivatives because i do not have any norm on $\mathcal{P}_n$ – user127.0.0.1 Dec 3 '12 at 16:26
I would reformulate the problem, using $\mathcal P_n\cong \Bbb R^{n+1}$ via the standard bases, i.e., $\mathfrak q$ is represented by the vector $(\mu_0,..,\mu_n)$ and then for $\mathfrak p=(\lambda_0,..\lambda_n)$, we'd have $f(\mathfrak p)=\sum_k \mu_k\left(\sum_j \lambda_j\right)^k$. – Berci Dec 3 '12 at 16:29
The standard basis of $\mathcal P_n$ is $1,x,x^2,..x^n$. – Berci Dec 3 '12 at 16:30
The question is actually even deeper - what do you mean by continous if you cannot measure distances in $\mathcal{P}_n$? – bonext Dec 3 '12 at 16:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.