Showing that the directional derivatives exist but f is not continuous

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  • #1
Hernaner28
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Homework Statement


It says:

[tex] \displaystyle f:{{\mathbb{R}}^{2}}\to \mathbb{R}[/tex]

[tex] \displaystyle f\left( x,y \right)=\left\{ \begin{align}
& 1\text{ if 0<y<}{{\text{x}}^{2}} \\
& 0\text{ in other cases} \\
\end{align} \right.[/tex]

Show that all the directional derivatives about (0,0) exist but f is not continuous in (0,0).

Homework Equations


Directional derivative:
[tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]


The Attempt at a Solution



I write the equation for the directionar derivative respect to a generic v=(a,b) about the origin:

[tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]

That is:

[tex] \displaystyle =\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( ha,hb \right)}{h}[/tex]

I should arrive that the limit does not depend on v=(a,b). But I'm stuck here. Which is the value of [tex] \displaystyle f\left( ha,hb \right)[/tex] 1 or 0? It clearly depends on the vector v. If I get close to the origin by the y-axis then [tex] \displaystyle f\left( 0,hb \right)=0[/tex].

Thanks!

PD: I hope I'm not breaking the rules. I've received several infractions which I'm not going to discuss here. It's not my intention.
 
Last edited:

Answers and Replies

  • #2
36,707
8,700

Homework Statement


It says:

[tex] \displaystyle f:{{\mathbb{R}}^{2}}\to \mathbb{R}[/tex]

[tex] \displaystyle f\left( x,y \right)=\left\{ \begin{align}
& 1\text{ if 0<y<}{{\text{x}}^{2}} \\
& 0\text{ in other cases} \\
\end{align} \right.[/tex]

Show that all the directional derivatives about (0,0) exist but f is not continuous in (0,0).

Homework Equations


Directional derivative:
[tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]


The Attempt at a Solution



I write the equation for the directionar derivative respect to a generic v=(a,b) about the origin:

[tex] \displaystyle \frac{\partial f}{\partial v}\left( 0,0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( \left( 0,0 \right)+\left( ha,hb \right) \right)-f\left( 0,0 \right)}{h}[/tex]

That is:

[tex] \displaystyle =\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( ha,hb \right)}{h}[/tex]

I should arrive that the limit does not depend on v=(a,b). But I'm stuck here. Which is the value of [tex] \displaystyle f\left( ha,hb \right)[/tex] 1 or 0? It clearly depends on the vector v. If I get close to the origin by the y-axis then [tex] \displaystyle f\left( 0,hb \right)=0[/tex].

Thanks!

PD: I hope I'm not breaking the rules. I've received several infractions which I'm not going to discuss here. It's not my intention.

Are you sure you have posted the exact wording of the problem? The function isn't continuous at (0, 0). In fact it's discontinuous all along the parabola y = x2 in the x-y plane.
 
  • #3
Hernaner28
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Yes, I can imagine what the graph looks like and its discontinuity is pretty obvious but what I'm asked to show is that the directional derivatives do exist around the origin but the function is still discontinuous.

Thanks Mark!
 
  • #4
36,707
8,700
I don't see how the directional derivative in the direction of -j, at (0, 0), could exist.

Look at the difference quotient using these values:

v = (0, -1), h = .1
v = (0, -1), h = .01
v = (0, -1), h = .001
and so on...
 
  • #5
Hernaner28
263
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I think you're right, that's why I made this thdread.

The directional derivative respecto to v=(0,-1) about (0,0) is:

[tex] \displaystyle \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 0,-h \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{h}[/tex]

which is infinite, isn't it?

Thanks!

PD: I am looking at the wording of the problem right now and it says that. Maybe just a mistake from the professors.
 
  • #6
Hernaner28
263
0
Sorry, that's wrong. The limit is zero in all directions because f(0,-h)=0. The numerator is zero, then the limit is zero.
 
  • #7
vela
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You should be able to prove that given (a,b), if h is small enough, f(ha,hb) will equal 0.
 
  • #8
Hernaner28
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Yes, I see. Do I have to do any formal proof with deltas and epsilon? Or can I just write it directly?

Thanks vela!
 
  • #9
vela
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Depends on what the grader wants. You'd know better than us.
 
  • #10
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You should be able to prove that given (a,b), if h is small enough, f(ha,hb) will equal 0.
What about my example, where (a, b) = (0, -1). No matter how small h is (but is still positive), f(ha, hb) = 1. Is there a mistake in my thinking here?
 
  • #11
vela
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f(x,y) = 0 when y≤0. It's only equal to 1 between the x-axis and the parabola y=x2.
 

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