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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.

We first prove a compact-square form of the Kolmogorov-Arnold representation theorem, including the construction needed for exact representation. We then remove the compactness restriction; a mere limiting sequence of two-variable approximations would not establish a finite expression using one-variable functions.

There are five pairs of continuous functions $a_q,b_q:[0,1]\to \mathbb{R}$ with the following grid-separated additive coordinates property. At every arbitrarily small mesh, choose for each $q$ a finite family of disjoint closed intervals in each coordinate. Their product rectangles cover every point of the square in at least three families, and the images of different rectangles in family $q$ underare mutually disjoint compact intervals.

Here is a proof of that existence assertion. Work in the complete metric space $X=C([0,1]{)}^{10}$ with its maximum uniform norm. For an integer $N\ge 1$, let $U_N$ comprise tuples having the preceding property with every interval shorter than $1/N$. Each $U_N$ is open: fix a witnessing finite collection of rectangles. Their compact image intervals have positive mutual separation within each family, so sufficiently small uniform perturbations of the ten functions preserve separation, while the domain cover is unchanged.

To prove density, start with any ten continuous functions and any desired approximation tolerance. Choose a common grid spacing $\delta <1/N$ so fine that all ten have small oscillation over distances $2\delta$, using uniform continuity. For the five families, use grid cuts shifted successively by $\delta /5$. Remove tiny open gaps about these cuts, of radius less than $\delta /20$. Gaps from different families are disjoint. The complementary components in $[0,1]$ are closed intervals shorter than $\delta$. Every coordinate belongs to an interval in at least four families, so a pair of coordinates belongs to product rectangles in at least three common families.

In each family, approximate $a_q$ and $b_q$ by constants on those intervals and interpolate linearly across the gaps. The constants can be chosen arbitrarily close to sampled original values, so the resulting functions approximate the originals uniformly. Perturb the finitely many constants slightly so that all sums $a_{q,I}+b_{q,J}$ for distinct pairs $(I,J)$ are different. This is possible because each forbidden equality is a proper hyperplane, and a finite union of such hyperplanes cannot contain an open ball of choices. The rectangle images are now distinct singleton sets. Thus $U_N$ is dense. The Baire category theorem gives a tuple in ${\bigcap }_{N\ge 1}{U}_N$. Fix that tuple once and for all.

Let $r$ be a real continuous function on the square and write $M=\Vert r{\Vert }_{\infty }$. If $M=0$ there is nothing to do. Otherwise choose a sufficiently fine witnessing cover that the oscillation of $r$ in every rectangle is at most $M/10$. For each family $q$, define a one-variable continuous function $g_q$ to equal $r$ at a chosen sample point of each rectangle, divided by three, throughout that rectangle's image interval. The image intervals are disjoint, so interpolate linearly between them and extend constantly beyond the extreme intervals. ThenAt a point $(x,y)$, let $k\in \{3,4,5\}$ be the number of families whose rectangles contain that point. In those $k$ families the corresponding summand differs from $r(x,y)/3$ by at most $M/30$. In the other families its absolute value is at most $M/3$. ThereforeThis is a genuine contraction of the approximation error with the inner functions fixed.

Starting with $r_0=F\in C([0,1{]}^2)$, repeat the construction and put $r_{j+1}=r_j-{\sum }_q{g}_{q,j}\circ s_q$. Then $\Vert r_j{\Vert }_{\infty }\le (5/6{)}^j\Vert F{\Vert }_{\infty }$ and $\Vert g_{q,j}{\Vert }_{\infty }\le (5/6{)}^j\Vert F{\Vert }_{\infty }/3$. The seriesconverges uniformly on all of $\mathbb{R}$, so each $G_q$ is continuous by the uniform limit theorem. Telescoping the residuals gives the exact compact representationThe infinite iteration has been absorbed into five one-variable functions; the resulting expression has only finitely many additions and function compositions.

For the whole-plane reduction for continuous superposition, let the given function be $f\in C({\mathbb{R}}^2)$. SetThis is finite on every bounded interval by compactness. Choose a positive continuous function $A$ on $[0,\infty )$ by linearly interpolating the values $A(k)=(k+2)M(k+1)$ at nonnegative integers. For $k\le t\le k+1$, monotonicity of the endpoint values gives $A(t)\ge A(k)\ge (1+t)M(t)$. Extend $A$ constantly to negative arguments. ConsequentlyLet $\kappa (x)=\frac{1}{2}+{\pi }^{-1}\arctan x$. Transport $b$ to $(0,1{)}^2$ with ${\kappa }^{-1}(s)=\tan (\pi (s-1/2))$. Extend the transported function by zero on the boundary of the square. The displayed bound proves continuity at every boundary point, including corners, because approaching the boundary forces at least one original coordinate to tend to infinity. Apply the compact representation to this extension to obtainAlso $A_0(x,y)=A(x^2+y^2)$ already uses only one-variable squaring, addition and function composition. Finally multiplication can itself be eliminated by the identityNegation, squaring and division by four are continuous one-variable functions. Hence the entire expression uses only the permitted one-variable continuous functions and addition, even for an unbounded $f$ on ${\mathbb{R}}^2$.

For $f\in C({\mathbb{R}}^2)$, choose a positive continuous function $A$ with $A(t)\ge (1+t){\max }_{x^2+y^2\le t}|f(x,y)|$ for $t\ge 0$. Such an $A$ is obtained by piecewise linear interpolation of a sufficiently large increasing sequence. Then $b(x,y)=f(x,y)/A(x^2+y^2)$ tends uniformly to zero at infinity. Transport $b$ to $(0,1{)}^2$ by $x=\tan (\pi (s-1/2))$ and extend it by zero on the boundary; the extension is continuous. Apply the compact Kolmogorov-Arnold representation theorem, then multiply by $A(x^2+y^2)$. Multiplication introduces no additional primitive operation because $uv=[(u+v{)}^2-(u-v{)}^2]/4$ uses only addition and continuous one-variable squaring and scaling.