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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!

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  • $\begingroup$ When you say "the function $x\mapsto f(x,y)$ is continuous", what exactly do you mean there? Are you considering $x$ as the identify function in some function space? Or do you mean that $f(\cdot,y)$ is continuous from $\Bbb{R}$ to $\Bbb{R}$? $\endgroup$ – dbx Aug 3 '18 at 18:36
  • $\begingroup$ @Minh did you find the two references I provided to be helpful? $\endgroup$ – Matt A Pelto Aug 6 '18 at 21:36
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    $\begingroup$ @MattAPelto You Reminisce about concept of Lipschitz continuous, but this condition in multiple variables function So hard to identify. $\endgroup$ – Minh Aug 7 '18 at 2:25
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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}

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    $\begingroup$ Oh, this problem use the definition of Lipschitz continuous. Could you show me some books to study Lipschitz continuous, please $\endgroup$ – Minh Aug 4 '18 at 11:26
  • $\begingroup$ Most real analysis textbooks mention Lipschitz continuous somewhere but I am unaware of a reference that discusses a family of functions such that every function in the family satisfies a common Lipschitz condition. I found 2 textbooks that discuss Lipschitz continuous more than the other real analysis textbooks that I am familiar (Rudin, Tao, Folland) with, (i) Royden: math.harvard.edu/~ctm/home/text/books/royden-fitzpatrick/… ; (ii) Bartle & Sherbert: sciencemathematicseducation.files.wordpress.com/2014/01/… $\endgroup$ – Matt A Pelto Aug 4 '18 at 19:38
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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 $$|f(x',y')-f(x,y)|\leq|f(x',y')-f(x',y)|+|f(x',y)-f(x,y)|\leq$$ $$\leq K|y'-y|+|f(x',y)-f(x,y)|<$$ $$<K|y'-y|+e/2\leq e.$$

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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 .

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  • $\begingroup$ Careful the fact that either $\{\,f(\cdot, y) \}_{y \in \mathbb{R}}$ is an equicontinuous family or else $\{\,f(x, \cdot) \}_{x \in \mathbb{R}}$ is an equicontinuous family seems important here. Define $p_1,p_2:\mathbb{R}^2 \to \mathbb{R}$ by $p_1(x,y)=x$ and $p_2(x,y)=y$, respectively. It is true that a function $f: \mathbb{R} \to \mathbb{R}^2$ is continuous if and only if both $p_1 \circ f$ and $p_2 \circ f$ are continuous. $\endgroup$ – Matt A Pelto Aug 4 '18 at 1:23
  • $\begingroup$ We would at least (at least as far as I know) need to additionally suppose that $f(K) \subset \mathbb{R}$ is compact whenever $K \subset \mathbb{R}^2$ is compact to prove that $f(\cdot, y)$ continuous for each $y \in \mathbb{R}$ and $f(x, \cdot)$ continuous for each $x \in \mathbb{R}$ implies that $f$ is continuous. Without this additional hypothesis that I mention, there are a few threads on here with counterexamples. $\endgroup$ – Matt A Pelto Aug 4 '18 at 1:36
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    $\begingroup$ @MattAPelto . Agreed. It is NOT true in general that continuity in each variable separately implies continuity on $\Bbb R^2$. $\endgroup$ – DanielWainfleet Aug 4 '18 at 2:45
  • $\begingroup$ I would like to mention that showing $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ continuous in each variable separately, and $f(K)$ compact whenever $K$ compact implies $f$ is continuous requires a proof by contradiction and the IVT. So in some constructive sense even this is a reaching claim. $\endgroup$ – Matt A Pelto Aug 4 '18 at 2:58
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    $\begingroup$ You're correct, thank you for pointing out my mistake. I`ve added a counter example to this claim. $\endgroup$ – Sar Aug 4 '18 at 8:10

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