Generalizing the Analyticity of holomorphic functions

Question

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that at each point in the plane, if $D$ is the derivative matrix (Jacobian) of $f$, then $D^TD=\lambda I$ for some $\lambda$ and $det(D)≥0$. Notice that if we imagine $f$ instead as a function from $\mathbb{C}\rightarrow\mathbb{C}$ by identifying $f(a,b)$ with $f(a+bi)$, then this condition is equivalent to asking for $f$ to be complex differentiable. Recall that if a function is holomorphic on its entire domain, then its analytic, and thus all higher order derivatives exist. Thus if this condition on the derivative matrix $D$ holds, then all higher derivatives of $f$ exist.

I would like to know if there exist generalizations of this fact to higher dimensions. For example, say $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ such that at each point $\mathbb{R}^n$, if $D$ is the derivative matrix (Jacobian) of $f$, then $D^TD=\lambda I$ for some $\lambda$ and $det(D)≥0$. Must all higher derivatives of $f$ exist in this case? If not, is there a different condition we can place on $D$ so that all higher derivatives of $f$ must exist?

In the case of looking for stronger conditions on $D$, of course we could pick conditions that trivially make all higher derivatives exist, for example, the condition that $D=I$ or $D=0$, however I would like to know about non trivial conditions if possible.

Answer 1

Here is the basic definition of Frechet-differentiability in Banach spaces.

Definition. Let $V,W$ be Banach spaces over the same field $\Bbb{F}\in\{\Bbb{R},\Bbb{C}\}$. Let $A\subset V$ be an open set, $a\in A$ and $f:A\to W$ be a given mapping. Suppose there is a (necessarily unique) $\Bbb{F}$-linear transformation $T:V\to W$ such that \begin{align} \lim_{h\to 0}\frac{\lVert f(a+h)-f(a)-T(h)\rVert}{\lVert h\rVert}&=0. \end{align} In the case where $\Bbb{F}=\Bbb{R}$, we shall say $f$ is (real) differentiable at $a$. In the case where $\Bbb{F}=\Bbb{C}$, we shall say $f$ is holomorphic at $a$. In either case, this unique $T$ is denoted by $Df_a$ (or any other symbol to this effect).

This definition encompasses all the possible cases. For example, we can take $V=\Bbb{C}^n,W=\Bbb{C}^m$ and we have a notion of holomorphicity in this case. Note that in the case where $V=\Bbb{C}$ ($W$ can be any complex Banach space, for instance $\Bbb{C}$), this is equivalent to the existence of the limit $f'(a):=\lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}$. In this case, the relationship between the two notions is that $f'(a)=Df_a(1)$.

Let us record the following almost trivial theorem.

Theorem. Let $V,W$ be complex Banach spaces, so we can also consider them as real Banach spaces in the obvious way. Let $A\subset V$ be open, $a\in A$ and $f:A\to W$ be a given mapping. Then, we have that $f$ is holomorphic at $a$ if and only if $f$ is real-differentiable at $a$, and its derivative in the real sense (which is a-priori only a real-linear transformation) is complex-linear. In this case, the derivatives in either case coincide (i.e they are the same transformation $Df_a:V\to W$)

This should be clear from the definition, because the only difference between the real and complex case is whether the linearity being considered is over $\Bbb{R}$ or over $\Bbb{C}$. Now, in your question, you're asking about a function defined over the reals, and asking about its holomorphicity. To be precise, we should first specify in what way we're identifying $\Bbb{R}^{2n}$ with a complex vector space like $\Bbb{C}^n$. For this reason, we note the following definition.

Definition. Let $V$ be a real vector space. A complex structure on $V$ means a real-linear transformation $J:V\to V$ such that $J^2=-\text{id}_V$.

Here are some simple facts:

• Given a complex structure $J$ on a real vector space $V$, by defining scalar multiplication as $(a+ib)v:= av + bJ(v)$, we can consider $V$ as a complex vector space (so application of $J$ defines the scalar multiplication by $i$).
• A finite-dimensional real vector space $V$ admits a complex structure if and only if $\dim_{\Bbb{R}}V$ is even.
• If $J$ is a complex structure on a finite-dimensional real vector space $V$ and $S:V\to V$ is a real-linear isomorphism, then $S\circ J\circ S^{-1}$ is also a complex structure on $V$. Conversely, given two complex structures $J,J'$ on a finite-dimensional real vector space, there exists a real-linear isomorphism $S:V\to V$ such that $J'=S\circ J \circ S^{-1}$.

So, let us now take $V=\Bbb{R}^{2n}$ and $W=\Bbb{R}^{2m}$. Fix complex structures $J$ on $V$ and $J'$ on $W$, so that we can consider $V,W$ as either real or complex Banach spaces. Let $A\subset V$ be open and $a\in A$ and $f:A\to W$ a given mapping. Then, by the trivial theorem above, we have that $f$ is holomorphic at $a$ if and only if $f$ is real-differentiable at $a$, and $Df_a$ is a complex-linear transformation. But now, being complex-linear is equivalent to the condition that $J'\circ Df_a= Df_a\circ J$ (i.e we can "pull out" scalar multiplication by $i$).

If we consider different complex structures, then the effect is to replace $f$ by an appropriate composition on the domain and target by real-linear isomorphisms.

For example, suppose $J:\Bbb{R}^{2n}\to\Bbb{R}^{2n}$ is defined by \begin{align} J(x_1,\dots, x_n,y_1,\dots, y_n):= (-y_1,\dots, -y_n,x_1,\dots, x_n), \end{align} and $J'$ defined similarly on $\Bbb{R}^{2m}$. Then, $f$ is holomorphic at $a$ (wrt complex vector space structure defined by $J,J'$) if and only if $f$ is real-differentiable at $a$ and the matrix representation (relative to standard ordered bases of $\Bbb{R}^{2n}$ and $\Bbb{R}^{2m}$) satisfies \begin{align} \begin{pmatrix} 0&-I_m\\ I_m&0 \end{pmatrix}\cdot [Df_a] &= [Df_a]\cdot \begin{pmatrix} 0&-I_n\\ I_n&0 \end{pmatrix}. \end{align}

If on the other hand we consider the complex structure defined by $M:\Bbb{R}^{2n}\to\Bbb{R}^{2n}$, \begin{align} M(x_1,y_1,\dots, x_n,y_n):= (-y_1,x_1,\dots, -y_n,x_n), \end{align} and likewise defining $M'$ on $\Bbb{R}^{2m}$, then $f$ is holomorphic at $a$ (wrt the complex vector space structure defined by $M,M'$) if and only if $f$ is real-differentiable at $a$ and \begin{align} \underbrace{\begin{pmatrix} J_1 & & \\ & \ddots &\\ & & J_1 \end{pmatrix}}_{\text{$m$ times }} \cdot [Df_a] &= [Df_a]\cdot \underbrace{\begin{pmatrix} J_1 & & \\ & \ddots &\\ & & J_1 \end{pmatrix}}_{\text{$n$ times}}, \end{align} where $J_1=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.

Now, having established the definitions, note that if we instead start with a function $f:A\subset \Bbb{C}^n\to\Bbb{C}^m$ then $f$ is holomorphic at $a$ if and only if each of its component functions $f_1,\dots, f_m$ is holomorphic at $a$ (according to the first definition). Thus, in what follows, we may as well assume $m=1$. The conditions above boil down to the appropriate Cauchy-Riemann equations \begin{align} \frac{\partial f}{\partial x_k}(a)+i\frac{\partial f}{\partial y_k}(a)&=0 \qquad \left(1\leq k \leq n\right). \end{align} (the different matrices arose above simply because of the manner in which we identified $\Bbb{R}^{2n}$ with $\Bbb{C}^n$).

So, now we can try to prove holomorphic on $A$ implies analytic on $A$ by mimicking the one-dimensional proof; use a generalized form of Cauchy's integral formula to write $f$ as a power series in $n$-variables (the coefficients being given by appropriate integrals). From here, several of the basic results can be generalized simply by going over the proofs in the one-dimensional case and modifying them appropriately.