Does anyone know whether the product/quotient rule for Fréchet derivatives still hold? For example, consider the evaluation operator: $$\rho_x : (C[a,b],\|\cdot\|_\infty) \rightarrow (\mathbb{R},|\cdot|)$$ where $\|\cdot\|_\infty$ is the sup-norm and $|\cdot|$ the Euclidean norm. Then I may define an operator: $T$ for $f\in C[a,b]$ acting as $$T(f) = \frac{\rho_x (f)}{\rho_y(f)} = \frac{f(x)}{f(y)}$$ (Assume the denominator is not zero). Knowing that the Fréchet derivative of $\rho_x$ is $\rho_x$ itself at any point $f\in C[a,b]$, what can we say about the Fréchet derivative of $T$?

Guess: $DT(f)(\cdot) = \frac{\rho_y(f)\rho_x(\cdot) + \rho_x(f)\rho_y(\cdot) }{\rho_y(f)^2} \in L(C[a,b],\mathbb{R})$

Thanks for you answers!


This is as simple as proving the quotient rule for real-valued functions. Assume $f(y) > 0$. \begin{align*}T(f+g) - T(f) - DT(f)\,g &= \frac{f(x) + g(x)}{f(y) + g(y)} - \frac{f(x)}{f(y)} - \frac{f(y) \, g(x) - f(x) \, g(y)}{f(y)^2}\\ &=\frac{f(x) \, f(y)^2 + g(x) \, f(y)^2 - f(x) \, f(y)^2 - f(x) \, f(y) \, g(y) - f(y)^2 \, g(x) - f(y) \, g(x) \, g(y) + f(x) \, g(y) \, f(y) + f(x) \, g(y)^2}{f(y)^2 \, (f(y) + g(y))}\\ &=\frac{- f(y) \, g(x) \, g(y) + f(x) \, g(y)^2}{f(y)^2 \, (f(y) + g(y))} \\ &= \mathcal{o}( \lVert g \rVert_\infty) \end{align*}

  • $\begingroup$ Yes, I did this as well, but wanted to be sure. Is this extendable to non-real-valued operators? For instance: $T:C^2:\rightarrow C$ with $T(f) = f'*f''$ ? $\endgroup$ – Mark Jun 14 '13 at 9:41
  • $\begingroup$ @Mark Yes, it is. Let $B \colon X \times Y \to Z$ be bilinear and continuous (for example the multiplication $f' \cdot f''$ on $X=Y=C^2$ with $Z= C$). Then for any differentiable $f \colon W \to X$, $g \colon W \to Y$, we have that $B(f,g) \colon W \to Z$ is differentiable with $D(B(f,g))(w)h = B(Df(w)h, g(w)) + B(f(w), Dg(w)h)$. $\endgroup$ – martini Jun 14 '13 at 11:47

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