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The definition is from this book https://link.springer.com/content/pdf/10.1007/978-0-387-68407-9.pdf?pdf=button

Fréchet derivatives in this book are just the usual higher dimension derivatives so I don’t write the definition here

Definition 1.20. Let $f$ from $U$ into $R^{m}$ be a map on an open set $U$ in $R^{n}$. We call $f$ twice Fréchet differentiable on $U$ if both $f$ and $Df$ are Fréchet differentiable on $U$, and denote by $D^{2}f := D(Df)$ the second derivative of $f$.

However, say $f$ is from $R^{n}$ into $R^{m}$, then doesn’t this mean at each point $x$, $Df(x)$ is an $m$ by $n$ matrix so $Df$ is a function from $R^{n}$ into the space of $m$ by $n$ matrices? Then what is $D(Df)$? Could someone explain a little? Thank you

(And I can’t understand the two equations in theorem 1.21 after “Since $Df$ is Fréchet differentiable” because it involves this definition 1.20 btw)

Tom
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    It could be what Jochen said. It can also just mean a matrix of the second partials. When people say a multivariate function is twice differentiable, they generally mean it’s $C^2$ – Andrew Jan 17 '23 at 17:32

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The derivative of $f:U\to\mathbb R^m$ is a function $Df: U\to L(\mathbb R^n,\mathbb R^m) \cong \mathbb R^{m\times n}$. The second derivative $D^2f=D(Df)$ is thus a function $U\to L(\mathbb R^n,L(\mathbb R^n,\mathbb R^m))$. This last space of linear $L(\mathbb R^n,\mathbb R^m)$-valued maps can be identified with the bilinear maps $\mathbb R^n\times\mathbb R^n\to \mathbb R^m$ via $f\mapsto B_f$ where $B_f(x,y)=f(x)(y)$. This identification is useful to formulate the symmetry of the second derivative.

Jochen
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  • Thanks a lot for the clarification! Also if you have time could you please take a look at this? It's something related https://math.stackexchange.com/questions/4620442/an-equation-about-the-2nd-order-fr%c3%a9chet-derivative – Tom Jan 17 '23 at 19:13