Let $f: U \subseteq \mathbb{C} \to \mathbb{C}^n$ be a vector-valued complex function of one variable. Is it true that $f$ is holomorphic if and only if any of its components is? And what about Cauchy-Riemann equations? Is it true that $f$ satisfies Cauchy-Riemann equations if and only if any of its component do satisfy it?
Cauchy-Riemann Equations for vector-valued complex function of one variable
complex-analysis
Best Answer
About weakly and strongly holomorphic vector-valued functions:
It is well-known that a vector-valued function $f : U \subset \mathbb{C} \to X$ with $X$ a Fréchet space (for example any Banach space is Fréchet) is strongly holomorphic, aka $\lim_{w \to z} \frac{f(z) - f(w)}{z - w}$ exists in $X$ for all $z \in U$, if and only if $f$ is weakly holomorphic, aka $\Lambda \circ f : U \to \mathbb{C}$ is holomorphic in the usual sense for all continuous linear functionals $\Lambda : X \to \mathbb{C}$ on $X$. You can define weak/strong continuity, weak/strong Lipschitz, etc... in the same fashion.
It is quite remarkable in the sense that the similar statement does not hold if we replace "holomorphic" by "continuous" in infinite-dimensional settings, you can for example look at the function $g : [0,1] \to \ell^2$ defined by: $$\begin{cases}g(0) := 0\\ g\left(\frac{1}{n}\right) = e^{(n)}\\ \text{$g$ affine between $\frac{1}{n+1}$ and $\frac{1}{n}$} \end{cases}$$ where $e^{(n)}$ is the sequence with only zeroes except for a $1$ at the $n$-th place.
This function is weakly continuous everywhere including at $0$ (to see that, you want to use the fact that for each $\Lambda$ there exists a sequence $u^{(\Lambda)} \in \ell^2$ such that $\Lambda : v \in \ell^2 \mapsto \sum_{n = 0}^{\infty} u^{(\Lambda)}_n v_n$, and then work with that expression) , and strongly continuous on $(0,1]$ (because it's piecwise affine), but it is not strongly continuous at $0$ (since that would imply that $\left(e^{(n)}\right)_n$ converges to $0$, which is false).
For a proof of the fact that weak holomorphicity implies strong holomorphicity (the other way round being easy to see by continuity of the $\Lambda$s), you can refer for example to Rudin's Functional Analysis, Chapter $3$: Convexity, and look at the parts about "Vector-Valued Integration" and "Holomorphic Functions" (well, especially the latter for the proof itself, but the former introduces the notion of integral that is used, which is easier to define than the Bochner integral and is still sufficient fo what we need. For continuous functions it should end up being the same as the Bochner integral anyway.).
A quick outline for $X$ a Banach space is this: you observe that a weakly holomorphic function is weakly Lipschitz on all closed discs in $U$, you then use Banach-Steinhaus on the functionals (acting on the dual, so they are from the dual of the dual): $$\Phi_{z,w} : \Lambda \in X^* \mapsto \frac{\Lambda(f(z)) - \Lambda(f(w))}{z - w}$$ for $(z,w) \in \overline{D(a,r)}^2, z \neq w$ to prove that a weakly Lipschitz function on a closed disc is strongly Lipschitz and thus strongly continuous on that disc.
Thus your weakly holomorphic function is strongly continuous on every closed disc in $U$ and as such everywhere in $U$ since it's open.
Afterwards, there's a few ways you can go: following Rudin you can show that $f$, being weakly holomorphic and strongly continuous, satisfies Cauchy's integral formula, and then prove that that means that $f$ is strongly holomorphic. But you can also show that on every open disc $f$ is the strong derivative of a strongly holomorphic function: $$F : z \mapsto \int_{[a,z]} f(\xi)\mathrm{d} \xi$$ (reasoning inspired by Rudin's Real and Complex Analysis, Chapter $10$: Elementary properties of holomorphic functions, theorems $10.14$ and $10.17$). Either way you go, you'll be able to prove that $f$ is strongly holomorphic.
The advantage of Rudin's way (the former one) is that it is self-contained in the sense that you go directly from weakly holomorphic to strongly holomorphic without external arguments, but the way I propose only relies on redeveloping/convincing yourself that you can redevelop the theory of strongly holomorphic functions the exact same way you do over $\mathbb{C}$, and then use the fact that strong derivatives of strongly holomorphic are themselves strongly holomorphic to conclude.
Back to the actual question:
By replacing $X$ by $\mathbb{C}^n$, which is indeed a Banach space, we are back to the question's context.
But now, we have a full description of the (continuous, but they all are continuous since it's finite-dimensional) linear functionals on $\mathbb{C}^n$: for each $\Lambda \in \left(\mathbb{C}^n\right)^*$, there exists a matrix $A^{(\Lambda)} =: \pmatrix{a^{(\Lambda)}_1 & \cdots & a^{(\Lambda)}_n} \in \mathcal{M}_{1,n}(\mathbb{C})$ such that: $$\forall \Lambda \in \left(\mathbb{C}^n\right)^*,\,\forall v \in \mathbb{C}^n,\quad \Lambda(v) = A^{(\Lambda)} v = \pmatrix{a^{(\Lambda)}_1 & \cdots & a^{(\Lambda)}_n} \pmatrix{v_1 \\ \vdots \\ v_n} = \sum_{k = 1}^{n} a^{(\Lambda)}_k v_k$$ What this means is that the $(\Lambda \circ f)$s are written like this, if we call $f$'s components $f_1, \dots, f_n$: $$\forall \Lambda \in \left(\mathbb{C}^n\right)^*,\,\forall z \in U,\quad \Lambda(f(z)) = \sum_{k = 1}^{n} a^{(\Lambda)}_k f_k(z)$$ If we denote by $\pi_k$ the $\Lambda$ of matrix $\pmatrix{0 & \cdots & 0 & 1 & 0 & \cdots & 0}$ with the $1$ at the $k$-th position, then it is clear that $f_k = \pi_k \circ f$ is holomorphic if $f$ is holomorphic. And if the $f_k$s are holomorphic, then by sum of holomorphic functions, $\Lambda \circ f = \sum_{k = 1}^n a^{(\Lambda)}_k \pi_k \circ f$ is holomorphic for all $\Lambda$s, and thus $f$ is holomorphic by what has been discussed previously.
This kind of reasoning might also be viable if instead of $\mathbb{C}^n$ we take a separable Hilbert space or an $\ell^p$ space ($p \neq \infty$) since you could define the same notion of "components" along each vector of an orthonormal basis/the Schauder basis of the $e^{(n)}$s, and the $\Lambda$s have the "same form", but I won't venture that far for now.
As for the Cauchy-Riemann equations: if we look at $f$ as a function $f(x,y)$ with $(x,y) \in \mathbb{R}^2$, and same for the components of $f$, then it should be simple to see that: $$\frac{\partial f}{\partial x}(x,y) = \pmatrix{\frac{\partial f_1}{\partial x}(x,y)\\ \vdots \\ \frac{\partial f_n}{\partial x}(x,y)} \quad\underline{\text{and}}\quad \frac{\partial f}{\partial y}(x,y) = \pmatrix{\frac{\partial f_1}{\partial y}(x,y)\\ \vdots \\ \frac{\partial f_n}{\partial y}(x,y)}$$ Therefore, if I denote the Cauchy-Riemann equations by $(\mathrm{CR})$ : $$\begin{split}\text{$f$ is holomorphic} &\Leftrightarrow \text{the $f_k$s are holomorphic}\\ &\Leftrightarrow \text{the $f_k$s are real-differentiable and satisfy $(\mathrm{CR})$}\\ &\Leftrightarrow \text{$f$ is real-differentiable and satisfies $(\mathrm{CR})$}\end{split}$$ You can prove the same thing when $f$ takes its values in an infinite-dimensional space without using "components of $f$" by directly proving the equivalence between holomorphicity and differentiability $+$ $(\mathrm{CR})$ using the notion of weak holomorphicity I talked about previously. The only trap is that, unlike here, you can't decompose a general vector into a "real part" and an "imaginary part" (here you could write $f = u + iv$ with $u, v : \mathbb{R} \to \mathbb{R}^n$), but that's not a problem if you stay with the Cauchy-Riemann equations stated like below: $$i \frac{\partial f}{\partial x}(z) = \frac{\partial f}{\partial y}(z)$$ after adapting the definitions of partial derivatives.
It is quite easy to see that for $\Lambda \in X^*$, $\Lambda \circ \frac{\partial f}{\partial x} = \frac{\partial (\Lambda \circ f)}{\partial x}$, going back to the definition of a derivative and the linearity of $\Lambda$, and the same goes for the other partial derivative, thus $f$ being a solution of the strong Cauchy-Riemann equations (on $f$) implies that it is a solution of the weak Cauchy-Riemann equations (on the $\Lambda \circ f$s). This also helps showing that if $f$ is strongly holomorphic, then $f$ satisfies the strong Cauchy-Riemann equations, by first noticing that it satisfies the weak Cauchy-Riemann equations.
But since there is already equivalence between weak Cauchy-Riemann $+$ real-differentiability and weak holomorphicity, and between weak holomorphicity and strong holomorphicity, these two implications close the loop, and so you get the equivalence between $f$ (strongly) holomorphic and $f$ being real-differentiable and satisfying the (strong) Cauchy-Riemann equations.