Bilinear interpolation

Remarks

The formula has the required form and takes the value $f_{ij}$ at each corner, which shows existence. Uniqueness holds because a function $a + bx + cy + dxy$ vanishing at the four corners has $a = b = c = d = 0$, as one checks on the unit square. Example: for $f(x, y) = 2 + 3x - y + xy$, the bilinear interpolant on any rectangle is $f$ itself, since $f$ already has the required form. Non-example: $p(x, y) = x^2$ on the unit square has corner values $0, 1, 0, 1$, but it is not of the form $a + bx + cy + dxy$, so it is not the bilinear interpolant; that interpolant is $x$. The interpolant is linear along every horizontal and vertical line, but not linear in general: along the diagonal $x = y$ of the unit square, $xy$ restricts to $x^2$. On a grid, the cellwise interpolants agree on shared edges, so the result is continuous, but its gradient jumps across cell edges in general.