If partial derivatives of $f$ is bounded, show that $f$ is continuous.

Question

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.

I edited now, $f_1$ is the derivative of $f$ with respect to $x$ and $f_2$ is with respect to $y$. — offret, Apr 11 '17 at 20:20
Hint: $f(p) - f(p_0) = f(p) - f(p_1, p_{0,1}) + f(p_1, p_{0,1}) - f(p_0)$. — user251257, Apr 11 '17 at 20:20
See also http://math.stackexchange.com/questions/351544/bounded-partial-derivatives-imply-continuity#412704 — user251257, Apr 11 '17 at 21:42

A-B-izi · Answer 1 · 2017-04-11T22:07:13.617

5

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.

edited Apr 11 '17 at 22:07

answered Apr 11 '17 at 21:10

A-B-izi

364

3

Your second sentence isn't what you want to say, is it? – zhw. Apr 11 '17 at 21:34
1

@zhw. Thank you, I edited my answer. I think now it should be correct. – A-B-izi Apr 11 '17 at 21:51
1

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. – zhw. Apr 11 '17 at 21:55
@zhw. You are right, boundedness is not sufficient; I just wanted a set where the supremum M is well defined. – A-B-izi Apr 11 '17 at 21:59
@A.Brizzi boundness is assumed ... – user251257 Apr 11 '17 at 22:00
@user251257 I am a bit tired as you can see :p – A-B-izi Apr 11 '17 at 22:02
@user251257 concerning the typos, could you be more explicit? – A-B-izi Apr 11 '17 at 22:04
@A.Brizzi I corrected it already. – user251257 Apr 11 '17 at 22:04
@user251257 I noticed it now, thank you again – A-B-izi Apr 11 '17 at 22:05
Why can we use the Mean Value Theorem here? I thought that it can be used only in the case of continuous functions which is exactly what we want to prove here. – zltn.guba Mar 23 '24 at 09:25

score 0 · Answer 2 · answered Apr 11 '17 at 21:13

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.

If partial derivatives of $f$ is bounded, show that $f$ is continuous.

2 Answers2

Linked