By Step 1, we have that for a fixed z that\left|\int_{[0,1]^{2}}(W-U)(x,y)W(x,z)W(y,z)dxdy\right|\le \|W-U\|_{\square}Therefore, when integrating over all z\in[0,1] we get that \left|\int_{[0,1]^{3}}(W-U)(x,y)W(x,z)W(y,z)dxdydz\right|\le \|W-U\|_{\square}Using this bound on each of the three summands, we get that the whole sum is bounded by 3\|W-U\|_{\square}.
