Let a function $f(x,y)$ be defined in an open set $D$ of the plane, and suppose that $f_1$ and $f_2$ are defined and bounded everywhere in $D$. Show that $f$ is continuous in $D$.

The answer says "Using the mean value theorem, show that $|f(p)-f(p_0)|\le M|p-p_0|$"

But in order to use the mean value theroem, shouldn't we assume f is a continuous function, which is the aim? How can we use it? Even if I use it, I couldn't quite get the statement answer is saying. Any help is welcomed. Thanks in advance.

  • $\begingroup$ What's $f_1$ and $f_2$? $\endgroup$ – Rutger Moody Apr 11 '17 at 20:18
  • $\begingroup$ I edited now, $f_1$ is the derivative of $f$ with respect to $x$ and $f_2$ is with respect to $y$. $\endgroup$ – offret Apr 11 '17 at 20:20
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    $\begingroup$ Hint: $f(p) - f(p_0) = f(p) - f(p_1, p_{0,1}) + f(p_1, p_{0,1}) - f(p_0)$. $\endgroup$ – user251257 Apr 11 '17 at 20:20
  • $\begingroup$ See also math.stackexchange.com/questions/351544/… $\endgroup$ – user251257 Apr 11 '17 at 21:42

Let $(x_0,y_0) \in D$ be arbitrary. Since $D$ is open, there exists a set $A=(a,b)\times (c,d)\subset D$ which contains $(x_0,y_0)$. Now, since $f_1$,$f_2$ exist everywhere and are bounded, $M=\sup_{(x,y) \in A}|f_1|+\sup_{(x,y) \in A}|f_2|$ is finite. So:

$|f(x_0,y_0)-f(x,y)|\leqslant|f(x_0,y_0)-f(x,y_0)|+|f(x,y_0)-f(x,y)| \leqslant M|x-x_0|+M|y-y_0|$

using MVT for $f_1$ and $f_2$.

I think this should work, let me know if you spot any mistakes or something is not clear.

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    $\begingroup$ Your second sentence isn't what you want to say, is it? $\endgroup$ – zhw. Apr 11 '17 at 21:34
  • $\begingroup$ @zhw. Thank you, I edited my answer. I think now it should be correct. $\endgroup$ – A-B-izi Apr 11 '17 at 21:51
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    $\begingroup$ It's still not right. After the first sentence you want to say something like " there is an open square $A$ centered at $(x_0,y_0)$ contained in $D.$ Then for any $(x,y)\in A, \dots $" Boundedness of $A $ is not the big deal, but having a set where you can project properly is. $\endgroup$ – zhw. Apr 11 '17 at 21:55
  • $\begingroup$ @zhw. You are right, boundedness is not sufficient; I just wanted a set where the supremum M is well defined. $\endgroup$ – A-B-izi Apr 11 '17 at 21:59
  • $\begingroup$ @A.Brizzi boundness is assumed ... $\endgroup$ – user251257 Apr 11 '17 at 22:00

Hint: Suppose we're in $(-1,1)^2$ and we want to prove continuity at $(0,0).$ Let $(x,y) \in (-1,1)^2.$ Then

$$f(x,y)-f(0,0) = f(x,y)-f(x,0) + f(x,0) - f(0,0).$$

Now $f$ is differentiable on each horizontal and vertical line segment in $(-1,1).$ That follows from the existence of partial derivatives. So the MVT will apply on any such segment.


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