设 $f:\bbR^2\to \bbR$ 为连续函数, 且满足条件 $$\bex f(x+1,y)=f(x,y+1)=f(x,y),\quad\forall\ (x,y)\in \bbR^2. \eex$$ 证明: $f$ 是一致连续函数.
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