Holomorphic Functions of Several Complex Variables — Examples

Writing $z_k=1+(z_k-1)$ for $k=0,1$ and expanding gives $z_0{z}_1=(1+(z_0-1))(1+(z_1-1))=1+(z_0-1)+(z_1-1)+(z_0-1)(z_1-1)$.

The right-hand side is a finite multi-indexed power series about $(1,1)$, with exactly the four nonzero coefficients stated above, so it converges absolutely on every bidisc centred at $(1,1)$.

The displayed finite series already equals $f$ everywhere, so it is in particular the power-series expansion of $f$ about $(1,1)$. Also ${\mathrm{\partial }}^{(1,0)}f(1,1)=1$, ${\mathrm{\partial }}^{(0,1)}f(1,1)=1$, and ${\mathrm{\partial }}^{(1,1)}f(1,1)=1$ by direct differentiation, while every derivative of order at least $2$ in one coordinate is $0$; these values match the displayed coefficients.

By [L1] and [L2], $\exp (z_0+z_1)=({\sum }_{m\ge 0}{z}_0^m\mathrm{/}m!)({\sum }_{n\ge 0}{z}_1^n\mathrm{/}n!)$.

Each one-variable exponential series is absolutely convergent. Moreover, for the double array $a_{mn}=z_0^m{z}_1^n\mathrm{/}(m!n!)$, [L2] gives ${\sum }_n\mathrm{\mid }{a}_{mn}\mathrm{\mid }=(\mathrm{\mid }{z}_0{\mathrm{\mid }}^m\mathrm{/}m!)\exp (\mathrm{\mid }{z}_1\mathrm{\mid })$ and then ${\sum }_m{\sum }_n\mathrm{\mid }{a}_{mn}\mathrm{\mid }=\exp (\mathrm{\mid }{z}_0\mathrm{\mid })\exp (\mathrm{\mid }{z}_1\mathrm{\mid })<\mathrm{\infty }$. Hence [L3] identifies the product in step 1.1 with the absolutely convergent diagonal enumeration ${\sum }_{N\ge 0}{\sum }_{m=0}^N{z}_0^m{z}_1^{N-m}\mathrm{/}(m!(N-m)!)$. The substitution $(m,n)=(m,N-m)$ enumerates every pair in ${\mathbb{N}}^2$ exactly once, and [L4] now permits regrouping the array in iterated order. Therefore $\exp (z_0+z_1)={\sum }_{m\ge 0}{\sum }_{n\ge 0}\frac{z_0^m{z}_1^n}{m!n!}={\sum }_{\alpha \in {\mathbb{N}}^2}\frac{z^{\alpha }}{\alpha !}\mathrm{.}$

On every bidisc the series is absolutely convergent because it is the product of two absolutely convergent one-variable exponential series there, and step 2.1 already identifies the coefficient of $z^{\alpha }$ as exactly $1\mathrm{/}\alpha !$.

If $\mathrm{\mid }{z}_1\mathrm{\mid }<1$, then [L1] gives $(1-z_1{)}^{-1}={\sum }_{j\ge 0}{z}_1^j$, so multiplying by $z_0$ yields $z_0\mathrm{/}(1-z_1)={\sum }_{j\ge 0}{z}_0{z}_1^j$.

In multi-index form this is the stated coefficient rule: the only monomials that appear are $z_0{z}_1^j$, so $c_{(1,j)}=1$ and every other coefficient is $0$.

The absolute-value series is ${\sum }_{j\ge 0}\mathrm{\mid }{z}_0\mathrm{\mid }\mathrm{\mid }{z}_1{\mathrm{\mid }}^j$. If $z_0\mathrm{\ne }0$, division by the positive constant $\mathrm{\mid }{z}_0\mathrm{\mid }$ and [L1] show that it converges exactly when $\mathrm{\mid }{z}_1\mathrm{\mid }<1$. If $z_0=0$, every term is $0$, so it converges for every $z_1$. Hence the absolute-convergence set is $(\mathbb{C}\times D(0,1))\cup (\{0\}\times \mathbb{C})$.

At points $(0,z_1)$ with $\mathrm{\mid }{z}_1\mathrm{\mid }\ge 1$, every term of the series is $0$, so the series still converges there to $0$, although the quotient is undefined when $z_1=1$ and this exceptional convergence set is not open.

Fix ${\zeta }_0$ on the circle $C$. As a function of ${\zeta }_1$, the integrand is ${\zeta }_0{\zeta }_1\mathrm{/}(({\zeta }_0-z_0)({\zeta }_1-z_1))={\zeta }_0({\zeta }_0-z_0{)}^{-1}\cdot {\zeta }_1({\zeta }_1-z_1{)}^{-1}$, where the scalar factor in front is constant in ${\zeta }_1$.

Apply [L2] to the holomorphic function ${\zeta }_1\mapsto {\zeta }_1$ on the disc $\mathrm{\mid }{\zeta }_1\mathrm{\mid }<1$: then $(2\pi i{)}^{-1}{\int }_C{\zeta }_1({\zeta }_1-z_1{)}^{-1}d{\zeta }_1=z_1$, so the inner integral equals $2\pi i\cdot {\zeta }_0{z}_1\mathrm{/}({\zeta }_0-z_0)$.

Substituting step 2.1 into the outer integral gives $\frac{1}{2\pi i}{\int }_C{\zeta }_0{z}_1({\zeta }_0-z_0{)}^{-1}d{\zeta }_0=z_1\cdot \frac{1}{2\pi i}{\int }_C{\zeta }_0({\zeta }_0-z_0{)}^{-1}d{\zeta }_0$, and [L2] applied again yields the value $z_1{z}_0$.

This matches the value $f(z)=z_0{z}_1$, exactly as [L1] predicts.

If $\mathrm{\mid }{\zeta }_0\mathrm{\mid }=\mathrm{\mid }{\zeta }_1\mathrm{\mid }=r$, then $\mathrm{\mid }f(\zeta )\mathrm{\mid }=\mathrm{\mid }\exp ({\zeta }_0+{\zeta }_1)\mathrm{\mid }=\exp (\mathrm{Re}{\zeta }_0+\mathrm{Re}{\zeta }_1)\le \exp (\mathrm{\mid }{\zeta }_0\mathrm{\mid }+\mathrm{\mid }{\zeta }_1\mathrm{\mid })=\exp (2r)$, with equality at ${\zeta }_0={\zeta }_1=r$. So ${\sup }_{{\mathrm{\Gamma }}_{(r,r)}(0)}\mathrm{\mid }f\mathrm{\mid }=\exp (2r)$.

Since [L2] makes $f$ holomorphic on every bidisc about $0$, applying [L1] with the supremum of step 1.1 gives $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }f(0)\mathrm{\mid }\le \alpha !\exp (2r)r^{-\mathrm{\mid }\alpha \mathrm{\mid }}$ for every multi-index $\alpha$.

By [L2], the coefficient of $z_0{z}_1$ is $1$, so [L4] gives ${\mathrm{\partial }}^{(1,1)}f(0)=1$. Thus the estimate for $\alpha =(1,1)$ is $1\le \exp (2r)\mathrm{/}{r}^2$.

The function $h(r)=\exp (2r)\mathrm{/}{r}^2$ satisfies $(\log h{)}^{\mathrm{\prime }}=2-2\mathrm{/}r$, so its unique critical point on $(0,\mathrm{\infty })$ is $r=1$, where it takes the value $e^2$; hence this example's best bound is $1\le e^2$.

The first component $(z_0,z_1)\mapsto z_0{z}_1$ and the second component $(z_0,z_1)\mapsto z_0+z_1$ are holomorphic by [L2], and the third component $(z_0,z_1)\mapsto \exp (z_0)$ is holomorphic by [L3].

Therefore [L1] makes $F$ holomorphic.

By [L4], $\mathrm{\partial }(z_0{z}_1)\mathrm{/}\mathrm{\partial }{z}_0=z_1$, $\mathrm{\partial }(z_0{z}_1)\mathrm{/}\mathrm{\partial }{z}_1=z_0$, $\mathrm{\partial }(z_0+z_1)\mathrm{/}\mathrm{\partial }{z}_0=\mathrm{\partial }(z_0+z_1)\mathrm{/}\mathrm{\partial }{z}_1=1$, $\mathrm{\partial }(\exp (z_0))\mathrm{/}\mathrm{\partial }{z}_0=\exp (z_0)$, and $\mathrm{\partial }(\exp (z_0))\mathrm{/}\mathrm{\partial }{z}_1=0$, which gives the displayed matrix.

Substituting $(z_0,z_1)=(0,0)$ into that matrix gives the displayed value at the origin.

By [L1], $J_{\mathbb{C}}F(z)$ is $\left(\begin{array}{cc}{z}_1& z_0\\ 1& 1\end{array}\right)$.

The swap $S$ is linear with matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, so $\det J_{\mathbb{C}}S=-1$ by [L2].

Apply [L2] to that matrix: $\det J_{\mathbb{C}}F(z)=z_1\cdot 1-z_0\cdot 1=z_1-z_0$, so the determinant vanishes exactly when $z_0=z_1$.

By [L3], $J_{\mathbb{C}}(S\circ F)=J_{\mathbb{C}}S(F(z))J_{\mathbb{C}}F(z)$, and [L4] gives $\det J_{\mathbb{C}}(S\circ F)=(-1)(z_1-z_0)=z_0-z_1$, exactly as the direct calculation of the swapped matrix would give.

For every $z$ in the closed unit bidisc, $\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\mid }{z}_0{z}_1\mathrm{\mid }=\mathrm{\mid }{z}_0\mathrm{\mid }\mathrm{\mid }{z}_1\mathrm{\mid }\le 1$, and equality holds if and only if $\mathrm{\mid }{z}_0\mathrm{\mid }=\mathrm{\mid }{z}_1\mathrm{\mid }=1$, that is, exactly on the distinguished boundary of [L2].

The point $(1,0)$ lies on the topological boundary of the closed unit bidisc but not on the distinguished boundary: every ball about $(1,0)$ meets the bidisc interior, while points with first coordinate of modulus $>1$ lie arbitrarily close outside it. At that point $\mathrm{\mid }f(1,0)\mathrm{\mid }=0$. So the whole topological boundary does not by itself identify where the maximum is attained.

This is exactly the concrete content of [L1] for the function $f(z)=z_0{z}_1$: the boundary points that matter are the distinguished ones.

The function $f(z_0,z_1)=z_0$ is holomorphic on ${\mathbb{C}}^2$ and is not identically zero, since $f(1,0)=1$.

Its zero set is exactly $\{(z_0,z_1)\in {\mathbb{C}}^2:z_0=0\}=\{0\}\times \mathbb{C}$.

This zero set is unbounded, because $(0,t)$ belongs to it for every complex $t$, and no point of it is isolated in it: if $(0,w)$ is a zero and $\epsilon >0$, then $(0,w+\epsilon \mathrm{/}2)$ is a different zero within Euclidean distance $\epsilon \mathrm{/}2$.

The zero set has empty interior, so this witness does not contradict [L2]; it shows instead that in several variables a nonzero holomorphic function can vanish on a whole positive-dimensional complex hyperplane. Therefore the refuted claim fails.

The false claim is the one-variable identity theorem [L1] repeated verbatim in two complex variables.

By [L3], the function $f(z_0,z_1)=z_0$ is holomorphic and not identically zero, while every point of its zero set $\{0\}\times \mathbb{C}$ is an accumulation point of that zero set. So the hypothesis of the false claim holds for this $f$.

Yet the conclusion fails, since $f(1,0)=1\mathrm{\ne }0$. What breaks from the one-variable proof is exactly [L2]: in one variable a nonzero holomorphic function has isolated zeros, whereas [L3] shows that in several variables a nonzero holomorphic function can vanish on a whole hyperplane. Therefore the false claim is false.

The false claim is the real-variable analogue of [L1], so refuting it shows that the several-variable holomorphic theorem is genuinely complex-analytic and not a formal separate-regularity statement.

The displayed formula defines a real-valued function on all of ${\mathbb{R}}^2$.

The function is bounded: for $(x,y)\mathrm{\ne }(0,0)$, the inequality $2\mathrm{\mid }xy\mathrm{\mid }\le x^2+y^2$ gives $\mathrm{\mid }f(x,y)\mathrm{\mid }\le 1\mathrm{/}2$, and at the origin $f(0,0)=0$, so $\mathrm{\mid }f(x,y)\mathrm{\mid }\le 1\mathrm{/}2$ everywhere.

The function is not continuous at $(0,0)$: if it were, then by [L3] applied with $\epsilon =1\mathrm{/}4$ there would be $\delta >0$ such that $\mathrm{\parallel }(x,y)\mathrm{\parallel }<\delta$ implies $\mathrm{\mid }f(x,y)\mathrm{\mid }<1\mathrm{/}4$; but for any real $t$ with $0<\mathrm{\mid }t\mathrm{\mid }<\delta \mathrm{/}\sqrt{2}$ one has $\mathrm{\parallel }(t,t)\mathrm{\parallel }=\sqrt{2}\mathrm{\mid }t\mathrm{\mid }<\delta$ and $f(t,t)=1\mathrm{/}2$, a contradiction.

Fix $b\mathrm{\ne }0$. Then the slice $x\mapsto f(x,b)=bx\mathrm{/}(x^2+b^2)$ is a quotient of polynomials, and the denominator never vanishes on $\mathbb{R}$, so [L2] makes this slice real analytic on all of $\mathbb{R}$. The same argument applies to the slice $y\mapsto f(a,y)$ when $a\mathrm{\ne }0$.

If $b=0$, then $f(x,0)=0$ for every real $x$, so the slice $x\mapsto f(x,0)$ is identically zero and therefore real analytic; by the symmetry $f(x,y)=f(y,x)$, the same is true of the slice $y\mapsto f(0,y)$. Thus every coordinate slice of $f$ is real analytic.

So this bounded function has real-analytic slices in each variable but is not jointly continuous, and the false claim fails. The published B-page item named cex-partial-derivatives-without-continuity uses the same witness for a different failure, but it is homed on an examples page and is therefore named here only in prose, without a wikilink or a dependency edge. The complex case is different precisely because [L4] turns a one-variable bound on a holomorphic slice into a uniform derivative bound, and no real-analytic hypothesis supplies such a bound.