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Suppose we have a function $f(x,y)$. Let it be a continuous one-to-one function.

If this function was evaluated at finite number of points $n$, $f(x_1,y_1),f(x_2,y_2),...f(x_n,y_n)$.

Can we conclude from the arguments and the value of the function at every point, what this function is? and if it's unique we can prove that it's the unique function to satisfy these constraints?

What about evaluating a function at infinite number of points(for example something of the form $f(x,b)=c$ for all $x$), can this uniquely specify it?

The motivation for this question is an answer I came up with in Physics stack exchange. And the constraints I imposed were the following

It's a continuous function.

$$\lim\limits_{y \mapsto \infty}f(x,y)=\infty$$ $$\lim\limits_{y \mapsto -\infty}f(x,y)=-\infty$$ $$f(x,x)=f(y,y)=0$$ $$f(x,0)=-x$$ $$f(0,y)=y$$

From this I concluded my function is $$f(x,y)=y-x$$

Does the above constraints uniquely specify $f(x,y)=y-x$, and is there a way to prove it?

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Omar Nagib
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    function $\langle x,y\rangle\mapsto y-x$ is not one-to-one. – drhab Jan 01 '16 at 14:21
  • @drhab Right, I removed that condition. – Omar Nagib Jan 01 '16 at 16:06
  • Your edit invalidates the specific answer but not the part that says it need not be $f(x,y)=y-x$. The new conditions only specify the function on the diagonals; perturb it anywhere else and it's still valid but not $y-x$. – Matt Samuel Jan 01 '16 at 16:39

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In fact both questions do address the same question what does that mean to "know" a function ? If you start from set-theoretical arguments a function is a graph that is a set $F$ (finite or not) of ordered pairs which is one-to-one i.e. if $\forall x$ there is at most only one pair $(x,y)$ in the graph with $x$ as the first coordinate. A function or a mapping is a the triple $(F,A,B)$ with $A$-set being the domain, i.e. the first projection of graph $F$, and $B$ any set larger than the range i.e. the second projection of graph $F$.

Knowing what a function is means knowing all the elements of the graph $F$ Either your data set covers all the graph F. Then you know exactly what is your function: this is $F=\{(x_n, f(x_n))\}$ in the finite case. Or this dataset corresponds to only sparsely sampled points picked up from the graph F, and such information is in general not sufficient to let you know what your function is unambiguously, unless some smoothness asumptions (e.g. $\mathcal C^0$,$\ldots$, $\mathcal C^\infty$) are formulated a priori about it. For example from Shannon's theory it is well known that if its Fourier transform has bounded support then the function can be recovered completely from an adequately sampled dataset (Shannon's sampling theorem). But in general you could imagine to fit the same discrete data set through interpolating polynomials of different orders or different families of polynomials, these polynomials being one-to-one once restricted to the support of your function.

All this discussion holds whether your consider $F$ as finite or non finite set. Obviously knowing only one partial application of a function of two variables, let say along $y=b$ is a cut made through the graph $F=\{((x,y),f(x,y))\}$ along one direction of the domain and is not sufficient to recover all the elements of the graph.

For instance the function $f(x,y) = -x + y(|x| + 1)$ satisfies your constraints,

$f(0,y)=y$ $f(x,0)=-x$ $\lim_{y\rightarrow\infty} f(x,y)=+\infty$ $\lim_{y\rightarrow-\infty} f(x,y)=-\infty$ is continuous.

and differs from the one you propose.

Hope this helps.

Pete
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Here is a picture of your function from wolfram alpha, obviously just a plane. There is nothing stopping me to deform this plane in a small area away from planes $ x = y,\ x = 0 \text{ and } y = 0 $ continuously and still satisfy all the conditions. Of course it's very hard to formulate such function explicitly and you have every right not to trust me until I provide a counter-example.

Now to a general question. A function, in a sense, is nothing but its graph: in a 2 variable case it's just a set $\{(x, y, f(x,y))|(x, y) \in D\}$ where D is the domain of f. Then obviously if you are given a finite set of points, you can do whatever you like inbetween, same for countably many points that don't form a dense set (see the comments). But because the function is continuous, if you are given uncountably many points, you may be forced to fill in the blanks in a special way (in particular if they form a dense subset - again see comments). Imagine you are given a set of points $\{(x,x^2)|x \in \mathbb{R} / \mathbb{Z} \}$. There is only one continuous function on $\mathbb{R}$ that satisfies all these points, $f(x) = x^2$, since any other assignment of f to integer values of x would result in a "spike".

This is just my intuition behind it, hopefully someone can formalize it.

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StasZ
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    $\mathbb{R} \setminus \mathbb{Z}$ is dense in $\mathbb{R}$, so any continuous function in uniquely determined by its values on $\mathbb{R} \setminus \mathbb{Z}$. But $\mathbb{Q}$ (which is countable) is also dense in $\mathbb{R}$, so a continuous function is also uniquely determined by its behaviour on $\mathbb{Q}$. – Jendrik Stelzner Jan 01 '16 at 14:54
  • Yes, my mistake. So a continuous function is uniquely defined by its values on a set A iff A is dense subset of the function's domain? Is there a name to this theorem? – StasZ Jan 01 '16 at 15:16
  • I don’t think this has a name; I’m also not sure when both implications hold. For Hausdorff topologocial spaces (e.g. metric spaces) it is true that a continuous function is uniquely determined by its restriction to a dense subset; in the case of a non-Hausdorff space this is not necessarily true. For the other implication I am not quite sure what we need: It holds at least for metric spaces, because for any subset $D$ the function $\mathrm{dist}(D,-)$ is $0$ on $\overline{D}$ but $>0$ on $(\overline{D})^C$. For more general topological spaces we probably need something like Urysohn or Tietze. – Jendrik Stelzner Jan 01 '16 at 15:30