prove function is constant

[SemperFi]

Hey everybody,

I want to show that a differentiable function f:R^n -> R which has the property <(grad f)(x),x>=0 for all x is constant.
Analogous to the theorem "f is constant, if f'=0" in the one-dimensional case, my approach to this problem would be
to use the mean value theorem (for multivariable functions). However, I don't make any progress.

Is there anyone who can help?

 

[Ishuda]

Hi everybody,

I want to show that a differentiable function f:R^n -> R that has the property <(grad f)(x),x>=0 for all x is constant.
Analogous to the proof of the theorem "f is constant, if f'=0" in the one-dimensional case, my approach to this problem
would be to use the mean value theorem (for multivariable functions), but it doesn't work somehow.

Is there anyone who could give me a hint?

In the multidimensional case, this can be done in sequence. Doing just for 2D let
grad(f) = (f_x, f_y)
we have, letting x = (x,y) = (1,0)
<grad(f), x> = f_x = 0
Thus, by the mean value theorem, f is independent of x, i.e.
f(x,y) = g(y).
Can you carry on from there?

 

[SemperFi]

In the multidimensional case, this can be done in sequence. Doing just for 2D let
grad(f) = (f_x, f_y)
we have, letting x = (x,y) = (1,0)
<grad(f), x> = f_x = 0

I don't understand this step. I mean, the values of the gradient depend on the point you choose, don't they?
So shouldn't it be <(grad f)(x),(x)>=<(grad f)(1,0), (1,0)>=f'_x(1,0)=0 then?

 

[Ishuda]

I don't understand this step. I mean, the values of the gradient depend on the point you choose, don't they?
So shouldn't it be <(grad f)(x),(x)>=<(grad f)(1,0), (1,0)>=f'_x(1,0)=0 then?

grad f = $\displaystyle \bigtriangledown\, f\, =\, (\frac{\partial\, f}{\partial\, x_1},\, \frac{\partial\, f}{\partial\, x_2},\, \frac{\partial\, f}{\partial\, x_3},\, ...,\, \frac{\partial\, f}{\partial\, x_n})$

I had assumed that the notation <a, b> was the inner (dot/scalar) product of two vectors a and b, i.e.
a = $\displaystyle (a_1,\, a_2,\, a_3,\, ...,\, a_n)$
and
b = $\displaystyle (b_1,\, b_2,\, b_3,\, ...,\, b_n)$
then
<a, b> = a₁ b₁ + a₂ b₂ + a₃ b₃ + ... + a_n b_n


EDIT: Yes grad f depends on what point you evaluate it at just as a derivative depends on the points you evaluate it at. However, grad f is a vector so it can not be equal to to the scalar zero. The dot product is a scalar and can not be equal to a vector.

 

[SemperFi]

grad f = $\displaystyle \bigtriangledown\, f\, =\, (\frac{\partial\, f}{\partial\, x_1},\, \frac{\partial\, f}{\partial\, x_2},\, \frac{\partial\, f}{\partial\, x_3},\, ...,\, \frac{\partial\, f}{\partial\, x_n})$

You are absolutely right with your assumptions. But still, shouldn't it be <(grad f)(x),x> = $\displaystyle \frac{\partial\, f}{\partial\, x_1}$(x)*x+$\displaystyle \frac{\partial\, f}{\partial\, x_2}$(x)*y = $\displaystyle \frac{\partial\, f}{\partial\, x_1}$(1,0)*1+$\displaystyle \frac{\partial\, f}{\partial\, x_2}$(1,0)*0 = $\displaystyle \frac{\partial\, f}{\partial\, x_1}$(1,0) ?

 

[SemperFi]

Ah, I see the misunderstanding: (grad f)(x) means "gradient evaluated at the point x" and not (grad f)*x

 

[Ishuda]

Ah, I see the misunderstanding: (grad f)(x) means "gradient evaluated at the point x" and not (grad f)*x

Yes