 Solution

We first prove the stronger compact-square form, in which all the one-variable functions are continuous. There are five fixed pairs of continuous functions $a_q,b_q$ on the unit interval such that every real $f\in C([0,1{]}^2)$ has a representationwith continuous one-variable $G_q$. We construct the inner functions and then prove convergence of the outer functions, rather than assume the Kolmogorov-Arnold representation theorem.

Constructing separating inner functions. Work in the Banach space $\mathcal{X}=C([0,1]{)}^{10}$ with the maximum of the ten supremum norms. For each positive integer $n$, let $U_n$ consist of tuples $(a_q,b_q{)}_{q=1}^5$ for which there are, for each $q$, finite families of closed intervals $I_{qi}$ and $J_{qj}$ of length less than $1/n$, satisfying two conditions. Each point of the unit interval belongs to at least four of the five unions ${\bigcup }_i{I}_{qi}$, and likewise for the $J$ families. Within each fixed $q$, the compact setsare pairwise disjoint for distinct pairs $(i,j)$. Each $K_{qij}$ is a closed interval, by continuity and the intermediate value theorem. These are grid-separated additive coordinates.

The set $U_n$ is open: for a tuple and its finitely many witnessing intervals, the distinct compact image intervals have a positive minimum separation. A sufficiently small uniform change in the inner functions preserves that separation, while the interval-cover conditions do not change.

To prove density, start with any ten continuous functions and any positive approximation tolerance. Choose a common fine partition so that each original function oscillates by less than a small fraction of the tolerance on a partition cell and its immediate neighbours; choose its mesh smaller than $1/(2n)$. Near every internal division point put five small, pairwise disjoint open gaps, one for each $q$. Use these gaps to divide the interval into closed towns for family $q$, including the two endpoints in towns. Gaps from different families are disjoint, so any point misses at most one family; every town has length less than $1/n$ if the gaps are sufficiently small. Do this for both coordinates.

Approximate $a_q$ and $b_q$ by functions which are constant on their respective towns, and interpolate linearly across the gaps. The towns lie near single partition cells, so uniform continuity makes these approximations as close to the original functions as desired. For each $q$, perturb the finitely many town values ${\alpha }_{qi},{\beta }_{qj}$ arbitrarily slightly so that all sums ${\alpha }_{qi}+{\beta }_{qj}$ are distinct. This is possible because each unwanted equality between distinct pairs is a proper affine hyperplane in the finite-dimensional vector space of values; finitely many such hyperplanes cannot fill any open ball. Interpolating the perturbed values still approximates the original functions, and the rectangle images are now distinct singleton sets. Thus $U_n$ is dense.

By the Baire category theorem, choose one tuple in ${\bigcap }_{n\ge 1}{U}_n$. To recall why completeness suffices here, start inside any open ball and successively choose closed balls of positive radii tending to zero, each contained in the preceding ball and in the next dense open $U_n$. Their centers form a Cauchy sequence, and its limit belongs to every ball, hence every $U_n$. Fix this tuple once and for all: it does not depend on $f$.

A quantitative approximation step. Let a continuous residual $r$ have supremum norm $M>0$. Choose $n$ large enough that the oscillation of $r$ on any rectangle of side lengths below $1/n$ is at most $M/20$. For each rectangle $I_{qi}\times J_{qj}$ choose a sample point $(x_{qi},y_{qj})$. Define a one-variable $h_q$ to have constant value $r(x_{qi},y_{qj})/3$ on $K_{qij}$. The image intervals are disjoint, so interpolate linearly in their gaps and extend constantly beyond the outermost ones. Then $h_q$ is continuous on the real line andAt a given $(x,y)$, at most one family fails to contain $x$ in an $I$ town and at most one fails to contain $y$ in a $J$ town. Thus the point belongs to rectangles in $k$ families, where $3\le k\le 5$. For each of these families the summand differs from $r(x,y)/3$ by at most $M/60$. Each remaining summand has absolute value at most $M/3$. ConsequentlyThis proves a strict contraction uniformly over the square, including points in gaps.

Summing the corrections. Set $r_0=f$ and apply that step repeatedly, subtracting the five summands at each stage. The residuals satisfy $\Vert r_m{\Vert }_{\infty }\le (3/4{)}^m\Vert f{\Vert }_{\infty }$, and the corresponding corrections satisfy $\Vert h_{qm}{\Vert }_{\infty }\le (3/4{)}^m\Vert f{\Vert }_{\infty }/3$. If a residual is zero, take all subsequent corrections to be zero. The seriesconverges uniformly on the real line to a continuous function. The residuals tend uniformly to zero, proving the boxed representation. An affine change of each coordinate treats any compact rectangle.

Removing a restriction to a compact domain. For a continuous real function $f$ on the whole plane, setChoose a positive continuous one-variable $A$ by linear interpolation of $A(n)=(n+2)(1+M(n+1))$ at nonnegative integers. Since these values increase, for $n\le t\le n+1$ we have $A(t)\ge (n+2)(1+M(n+1))\ge (1+t)(1+M(t))$. ThereforeLet $\sigma (x)=\frac{1}{2}+{\pi }^{-1}\arctan x$. Transport $b$ to the open square by $\tilde{b}(s,t)=b(\tan (\pi (s-1/2)),\tan (\pi (t-1/2)))$, and set it to zero on the square's boundary. The displayed decay estimate proves continuity at every boundary point, including the corners. Apply the compact-square representation to $\tilde{b}$ and substitute $s=\sigma (x)$, $t=\sigma (y)$ to obtain $b$ using continuous one-variable functions and addition.

Finally multiply this expression by $A(x^2+y^2)$. Multiplication itself uses only addition and one-variable functions, becauseSquaring, negation and scaling are one-variable operations, as are $A$ and $\sigma$. Thus the whole-plane reduction for continuous superposition gives a finite expression of the required kind for every continuous $f$ on the plane, without a boundedness assumption. For a complex-valued $f$, apply the argument to its real and imaginary parts and combine them using addition and the one-variable map $z\mapsto iz$. All the constituent one-variable functions can be taken continuous.