Show that $f$ is continuous on $\mathbb{R}^2$

Question

Consider function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. Suppose $\forall y \in \mathbb{R}$, the function $f(\cdot,y)$ is continuous and there is exists a positive number $K$ such that

$$|f(x,y') - f(x,y'')| \leq K |y' - y''|, \forall x,y',y'' \in \mathbb{R}$$

Show that $f$ is continuous on $\mathbb{R}^2$.

I have no idea to solve this problem, so I need help. Thanks all!

Answer 1

Let $(a, b) \in \mathbb{R}^2$, and let $\varepsilon>0$ be given. Since $f(\cdot,b)$ is continuous, we may find $\delta_1>0$ so that \begin{equation}\left|\, f(x,b)-f(a,b) \right|<\frac{\varepsilon}{2} \text{ whenever } |x-a|<\delta_1. \end{equation} Since $K>0$ is a common Lipschitz constant for the family of functions $\mathscr{F}=\{\,f(x, \cdot) \}_{x \in \mathbb{R}}$, we know that \begin{equation}\left|\, f(x,y)-f(x,b) \right|<\frac{\varepsilon}{2} \text{ for all } x \in \mathbb{R} \text{ whenever } K|y-b|<\frac{\varepsilon}{2}. \end{equation} So we select $\delta_2=\min\left\{\frac{\varepsilon}{2},\frac{\varepsilon}{2K}\right\}$. Note that for $(x,y),(a,b)\in \mathbb{R}^2$ we have $|x-a|+|y-b|\leq \sqrt{2}\sqrt{(x-a)^2+(y-b)^2}$. Setting $\delta=\frac{1}{\sqrt{2}}\min\{\delta_1, \delta_2\}$ it now follows that \begin{aligned}\left|\, f(x,y)-f(a,b) \right| &\leq \left|\, f(x,y)-f(x,b) \right| + \left|\, f(x,b)-f(a,b) \right|\\&<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon \, \text{ whenever }\, \sqrt{(x-a)^2+(y-b)^2}<\delta. \end{aligned}

Answer 2

Given $f(\cdot,y)$ is continous, that is , for any fixed $y$, $f(x)=f(x,y)$ is continous .We only need to show $f(x,\cdot$) is continous (A function is continous iff its continous for any of its variables).

Lets take $y'=y$ and $y''=y+h$, we have : $|f(x,y)-f(x,y+h)|\leq Kh \to 0$ meaning $lim_{h\to 0}f(x,y+h)=f(x,y)$,

and from here , $f$ is continous.

Edit: This answer is incorrect, A counter exaple to the claim "$f$ is contninous iff it is continous on every variable" would be : $$ g(x,y)=\begin{cases}{\frac {xy} {x^2+y^2}} \\0 \space\space\space\space\space, (x,y)=(0,0)\end{cases}$$ Fixing x or y we get a continous function at $0$, however, looking at the path $(t,t)$ we can see that $g(x,y)$ isnt continous at $0$.

However,similarly to other answers, we can write $|f(x,y)-f(a,b)|\leq |f(x,y)-f(x,b)|+|f(x,b)-f(a,b)|\leq K|y-b|+|f(x,b)-f(a,b)|$

From here, we can take $\delta$ s.t y is sufficiently closed to b and that $|f(x,b)-f(a,b)|<\epsilon/2$ from continouity .

Answer 3

Given $(x,y)$ and given $e>0$ let $d>0$ where $\forall x'\;(|x'-x|<d \implies |f(x',y)-f(x,y)|<e/2).$ We know $d$ exists because $f(\cdot,y)$ is continuous.

Let $d'=\min (d, e/2K).$ Then whenever $|x'-x|<d'$ and $|y'-y|<d'$ we have $$|f(x',y')-f(x,y)|\leq|f(x',y')-f(x',y)|+|f(x',y)-f(x,y)|\leq$$ $$\leq K|y'-y|+|f(x',y)-f(x,y)|<$$ $$<K|y'-y|+e/2\leq e.$$