Frechet derivative notation question

I am working with the Frechet derivative in my real analysis class and I have a question concerning its notation.

We have that for $V,W$ normed vector spaces, $A_{open} \subset V, f:A \rightarrow W, x_0 \in A,$ $f$ is Frechet differentiable at iff

$$\exists Df(x): V \rightarrow W \,s.t. \; \lim_{x\rightarrow x_0} \frac{||f(x) - f(x_0) - Df(x)(x-x_0)||_W} {||x-x_0||_V} = 0,$$ where $(x-x_0)$ is a perturbation of $x$. My confusion is coming from the notation $Df(x)(x-x_0)$. I know this is the Frechet derivative, but in terms of the the function itself, is $Df(x)(x-x_0)$ being evaluated at x, or at the perturbation?

• $Df$ is evaluated at $x$. That gives you a linear operator $Df(x)\colon V \to W$. That operator is then applied to $x-x_0$. – Daniel Fischer Dec 5 '13 at 17:09
• Ok. So $Df$ is evaluated at $x \in V$. Let's say that $f:R^n \rightarrow R, f(x) = x^TAx$ where $x \in R^n$ and $A \in R^{nxn}.$ I found that $Df(x)(v) = v^TAx + x^TAv,$ where $v$ is a perturbation. What would Df(x) look like in this case? Intuitively it seems that it would have the form $Df(x) = x^TAx + x^TAx$, but then the question becomes how do you apply $Df(x)$ to $v$. – DivDiff Dec 5 '13 at 18:15
• $Df(x)$ would be the map $v \mapsto v^TAx + x^TAv$. In components, you could express that as $Df(x) = x^T(A + A^T)$ here. – Daniel Fischer Dec 5 '13 at 18:26
• Ok. I'm having some difficulty seeing why that's the case. I see that $x^T(A + A^T) = x^TA + x^TA^T = x^TA + (Ax)^T,$ however I'm not seeing the connection between this and $Df(x)(v)$. – DivDiff Dec 5 '13 at 18:34
• $v^TAx + x^TAv = \left(v^TAx\right)^T + x^TAv = x^TA^Tv + x^TAv = (x^TA^T + x^TA)v = \bigl(x^T(A^T+A)\bigr)v$. So the mapping $v \mapsto v^TAx + x^TAv$ is given by the matrix $x^T(A^T+A)$. – Daniel Fischer Dec 5 '13 at 18:37

The formula isn't right: there is some confusion between $x$ and $x_0$. Correct version: $f$ is Frechét differentiable at $x_0$ iff $$\exists Df(x_0): V \rightarrow W \,\text{ s.t. }\; \lim_{x\rightarrow x_0} \frac{||f(x) - f(x_0) - Df(x_0)(x-x_0)||_W} {||x-x_0||_V} = 0$$
But I would rather present this in a slightly different language: $f$ is Frechét differentiable at $x_0$ iff there exists a bounded linear operator $T:V\to W$ such that $$\lim_{x\rightarrow x_0} \frac{||f(x) - f(x_0) - T(x-x_0)||_W} {||x-x_0||_V} = 0$$ If this holds, we call $T$ the Frechét derivative of $f$ at $x_0$, and denote it by $Df(x_0)$.