If f ( x , y ) f(x,y) , a 2 π 2\pi periodic function in each variable, has a modulus of continuity w f ( δ ) = o ( 1 / log ( 1 / δ ) ) {w_f}(\delta ) = o(1/\log (1/\delta )) then \[ σ ~ n ( x , y , f ) − ∫ 1 / n π ∫ 1 / n π [ f ( x + u , y + v ) − f ( x − u , y + v ) − f ( x + u , y − v ) + f ( x − u , y − v ) ] 4 tan ( u / 2 ) tan ( v / 2 ) d u d v → 0 uniformly in ( x , y ) {\tilde \sigma _n}(x,y,f) - \int _{1/n}^\pi {\int _{1/n}^\pi {\frac {{[f(x + u,y + v) - f(x - u,y + v) - f(x + u,y - v) + f(x - u,y - v)]}}{{4\tan (u/2)\tan (v/2)}}} } dudv \to 0\quad {\text {uniformly}}\;{\text {in}}\;(x,y) \] where σ ~ n ( x , y , f ) {\tilde \sigma _n}(x,y,f) is the first arithmetic mean of the conjugate series. This theorem is best possible in that o ( 1 / log ( 1 / δ ) ) o(1/\log (1/\delta )) cannot be replaced by O ( 1 / log ( 1 / δ ) ) O(1/\log (1/\delta )) .