Let $f(x,y)$ be continuous in $x$ when fixing $y$, and it is also continuous in $y$ when fixing $x$. Is $f$ continuous on its domain?
It seems very simple and straight-forward but could you please give me some hints to get started?
I am trying something like this: suppose $f$ is not continuous, then for some sequence $(x_n,y_n)\to (x,y)$, $\lim_{n\to \infty}f(x_n,y_n)\neq f(x,y)$. Let $\epsilon> 0$, wlog we assume $f(x_n,y_n)>f(x,y)+\epsilon$. However, by definition, $f(x_n,y_n)<f(x_n,y)+\delta$ for any $\delta$, and $f(x_n,y)<f(x,y)+\delta$ for any $\delta$. Contradiction!
I think this proof might only work for monotonic increasing (or decreasing) $f$. Not sure if this is correct.
It is not true. Take$$\begin{array}{rccc}f\colon&\mathbb R^2&\longrightarrow&\mathbb R\\&(x,y)&\mapsto&\begin{cases}\dfrac{xy}{x^2+y^2}&\text{ if }(x,y)\neq(0,0)\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $f$ is not continuous (since $\lim_{x\to0}f(x,x)=\frac12\neq f(0,0)$), but, for each $a\in\mathbb R$, the maps $x\mapsto f(x,a)$ and $x\mapsto f(a,x)$ are continuous.