### 35th Putnam 1974

Problem B4

R is the reals. f : R2→R is such that fx0 : R → R defined by fx0(x) = f(x0, x) is continuous for every x0 and gy0 : R → R defined by gy0(x) = f(x, y0) is continuous for every y0. Show that there is a sequence of continuous functions hn : R2 → R which tend to f pointwise.

Solution

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

We show first that hn tends to f pointwise. Take any point P in R2. 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 Mn flanking P and closer than δ to it. Let these lines cut L at Q and R. Then hn(P) lies between the values of f(Q) and f(R) and hence within ε of f(P).

It remains to show that hn is continuous. This is slightly tricky to nail down rigorously.Take any point P = (x0, y0) and any ε > 0. We need to show that hn takes poins close to P to values close to hn(P). Let L be the horizontal line through P. Assume that P does not lie on a line of Mn, 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, hn takes points inside the rectangle |x - x0| < δ ' , |y - y0| < δ to values within ε of hn(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 hn. 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 hn(x, y), where |y - y0| < 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 y0 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 hn(x, y) for |x - x0| < δ ' and |y - y0| < δ is less than 2ε, which is the statement that hn is continuous.

Finally, note that if P lies on a line of Mn, 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 hn is continuous. But having said that, my argument does seem rather complicated. Can anyone see a simpler argument? Or indeed a better choice of hn to make the whole argument simpler? (2) Note that for the particular hn we have chosen, tending to f pointwise needs f(x, y) continuous in x, whilst continuity needs f(x, y) continuous in y.