What can be said about $f$?

Question

I faced this question in an interview yesterday.

QUESTION: We have a function $f$ depending on three variables $x_1,x_2,x_3$. Now the gradient of $f$ is perpendicular at any point $(x_1,x_2,x_3)$, or in mathematical language , $$\vec r \cdot \nabla f =0$$ Now what can be said about $f$ from this information?

I could not make out anything from this. And I failed to proceed since I did not know the relation between $x_1,x_2,x_3$. I was given the hint that I should consider the function $f(t\vec r)$ where $t$ is a scalar and differentiate it with respect to $t$. I calculated it as: $$\frac{d}{dt}\left[f(t\vec r)\right]=f'(t\vec r) \vec r$$

I was still unable to make any interpretation. I was asked what the $f'$ thing meant. I said that it was a derivative but he asked me what it signified. But I had no clue.

Can anybody help me?

P.S. You can take $\vec r=(x_1,x_2,x_3)$. The interviewer said nothing about this. So I think it will be that way only.

Answer 1

Your calculations are wrong. It should read as

$$\frac d{dt} f(tx_1, tx_2, tx_3) = \frac{ \partial f}{\partial x_1} \frac { \partial (tx_1)}{\partial t} + \frac{ \partial f}{\partial x_2} \frac { \partial (tx_2)}{\partial t} + \frac{ \partial f}{\partial x_3} \frac { \partial (tx_3)}{\partial t} = \nabla f \cdot \vec r = 0$$

that is, if you fix a direction (fix $x_1, x_2, x_3$) your function is constant along that direction.

Hence, your function only depends on the angle (or the direction) of the point, not on the distance. That is

$$f(\vec r) = f(\vec r/a)$$

for every $a > 0$

Note that in general this is not constant.

It is constant if it is continuous at the origin: in this case, for every point $(x_1, x_2, x_3)$ you can "go back" along the line connecting this point to $0$ (without changing value); and if $f$ is continuous at $0$, then it must be that $f(x_1, x_2, x_3) = f(0,0,0)$ for any $(x_1, x_2, x_3)$.

On the other hand if $f$ is not continuous (or not defined) in $0$, then you will have some sort of "sphere" that generates all the values of the function, and they can be different

Answer 2

Update: My old proof had an error.

Now the gradient of $f$ is perpendicular at any point $(x_1,x_2,x_3)$,

$$ \DeclareMathOperator{grad}{grad} 0 = \grad f \cdot x = x_1 \partial_1 f + x_2 \partial_2 f + x_3 \partial_3 f \quad (1) $$

This works if the gradient is zero, which would mean $f$ is constant.

But can there be a non-zero gradient?

Below is an image of the 2D case $h(x_1, x_2) = (x_1, x_2)$ (red arrows) and a possible orthogonal vector field (green arrows) $g(x_1, x_2) = (-x_2, x_1)$, both arrow lengths reduced to $15\%$.

(Large Version, Source)

Problem: This field $g$ seems to have a non-zero curl and can not be a gradient field, as gradient fields have zero curl.

I would love to have a proof of this, that non-zero orthogonal vector fields to $h$ can not be gradient fields, but I have not found one yet.

The other argument is using the level curves:

The gradient is orthogonal to the level curves $f = \text{const}$, which would have the vectors $h(x) = x = r e_r$ as tangent vectors. So we would have curves $$ r(t) = m t \quad (t \in \mathbb{R}) $$ for constant vectors $m$ where $f$ is constant.

However all those curves of same level intersect at the origin, so they must share the same value.

BTW the term $\frac{d}{dt}f(t x)$ at $t=0$ is the directional derivative of $f$ at the origin in $x$-direction $$ D_x f(0) = f'_{0,x} = \left. \frac{d}{dt}f(0 + t x) \right\vert_{t=0} = \left. \grad f \right\vert_0 \cdot x = \left. \frac{\partial f}{\partial x} \right\vert_0= 0 $$