R is the reals. f **:** R^{2}→R is such that f_{x0} **:** R → R defined by f_{x0}(x) = f(x_{0}, x) is continuous for every x_{0} and g_{y0} **:** R → R defined by g_{y0}(x) = f(x, y_{0}) is continuous for every y_{0}. Show that there is a sequence of continuous functions h_{n} **:** R^{2} → R which tend to f pointwise.

**Solution**

Let M_{n} be the set of vertical lines at 1/2_{n} spacing, including the y-axis, ie all lines y = m/2^{n} for integral m. Define h_{n} to agree with f on M_{n} and to be horizontally interpolated elsewhere, so if x = m/2^{n} + x' , where 0 ≤ x' < 1/2^{n}, then h_{n}(x, y) = h_{n}(x, y) + (x'/2^{n})(h_{n}(x + 1/2^{n}, y) - h_{n}(x, y) ).

We show first that h_{n} tends to f pointwise. Take any point P in R^{2}. f is continuous along the horizontal line L through P, so given any ε > 0, we may take δ > 0 so that points on L within δ of P are taken by f to points within ε of f(P). Take n sufficiently large so that the there are adjacent lines of M_{n} flanking P and closer than δ to it. Let these lines cut L at Q and R. Then h_{n}(P) lies between the values of f(Q) and f(R) and hence within ε of f(P).

It remains to show that h_{n} is continuous. This is slightly tricky to nail down rigorously.Take any point P = (x_{0}, y_{0}) and any ε > 0. We need to show that h_{n} takes poins close to P to values close to h_{n}(P). Let L be the horizontal line through P. Assume that P does not lie on a line of M_{n}, so take H to be the first line to the left of P, and K the first to the right of P. Let H meet L at Q, and K meet L at R. Note that the separation of Q and R is non-zero and fixed, so the values f(Q) and f(R) may differ by a large amount. But we may take δ > 0 so that points on H within δ of Q are taken by f to values within ε/2 of f(Q) and points on K within δ of R are taken by f to values within ε/2 of f(R). We want to show that for sufficiently small δ ' > 0, h_{n} takes points inside the rectangle |x - x_{0}| < δ ' , |y - y_{0}| < δ to values within ε of h_{n}(P).

It is now easiest to think geometrically. Let us retain the x-axis to represent values of x, but now use the y-axis to represent possible values of h_{n}. So at Q we have a vertical bar height ε centred on f(Q) and at R we have another vertical bar height ε centred on f(R). Joining the tops of these bars and joining the bottoms gives a parallelogram. A vertical line through this parallelogram at x gives the possible values of h_{n}(x, y), where |y - y_{0}| < δ (because to get that value we must interpolate linearly between some point on the left-hand edge of the parallelogram and some point on the right-hand edge). Now suppose the top and bottom edges of the parallelogram make an angle θ with the y-axis. Then a thin vertical slice of the parallelogram centred on y_{0} width δ ' will project onto a segment length ε + δ ' cot θ. By taking δ ' sufficiently small we can make this less than 2ε. In other words the spread of values of h_{n}(x, y) for |x - x_{0}| < δ ' and |y - y_{0}| < δ is less than 2ε, which is the statement that h_{n} is continuous.

Finally, note that if P lies on a line of M_{n}, then we can deal separately with points to the left and right of P, using the same argument.

*Comments. (1) It is easy to botch the argument that h _{n} is continuous. But having said that, my argument does seem rather complicated. Can anyone see a simpler argument? Or indeed a better choice of h_{n} to make the whole argument simpler? (2) Note that for the particular h_{n} we have chosen, tending to f pointwise needs f(x, y) continuous in x, whilst continuity needs f(x, y) continuous in y. *

© John Scholes

jscholes@kalva.demon.co.uk

18 Aug 2001