# Critical point of a function - $\Bbb R^n$ Analysis

Consider $f=(f_1,f_2,f_3): U \rightarrow \mathbb{R}^3$ a function not identically null, $f\in C^1$ and rank $3$ at every point of the open $U \subset \mathbb{R}^n$, $n \geq 3$. Show that $g(x)= f_1^2(x)+f_2^2(x)+f_3^2(x)$, $x \in U$, hasn't maximum in $U$.

Suggestion: Suppose by contradiction considering $\nabla g$; look the sign of $g$.

• I edited your question for spelling and grammar; and $\LaTeX$ified the title. Hope the results are satisfactory and your meaning, mathematically, was preserved. Remember you can always roll back the version if I've erred, or let me know and I'll tend to it! Cheers! – Robert Lewis Jun 7 '15 at 19:39
• I also fixed the equation $g(x) = \text{etc.}$ – Robert Lewis Jun 7 '15 at 19:45
• @RobertLewis Ok, I appreciate. Thanks. – Let DC Jun 7 '15 at 19:47

There is a 'geometry' intuition behind your question. Observe that $g$ is the composition of $f$ and the square of the distance to the origen in $\mathbb{R}^3$. Namely, let $s : \mathbb{R}^3 \to \mathbb{R}$ be $s(x,y,z) := x^2 + y^2 + z^2$ be the square of the distance to $(0,0,0)$. Then $g = s \circ f$. Since $f$ has rank 3 at any point $x_0$ then there are vectors $v$ at $x_0$ such that the derivative of $f$ at $x_0$ takes $v$ to a vector $df(v)$ at $f(x_0)$ pointing outside of the sphere whose radius is $g(x_0)$. So $x_0$ can not be a maximum of $g$.

• How can I justify that $f(x_0)$ is outside of the sphere? – Let DC Jun 7 '15 at 19:14
• The vector $df(v)$ points outside the sphere of radius $g(x_0)$. – Holonomia Jun 7 '15 at 19:16
• Oh, ok, sorry. But why? – Let DC Jun 7 '15 at 19:17
• Because having rank 3 means that the derivative map is surjective. So you are free to choose the vector $w$ pointing outside the sphere of radius $g(x_0)$ and you are sure that there is a $v$ such that $df(v) = w$. – Holonomia Jun 7 '15 at 19:19
• And how there vectors $v$ at $x_0$ such that $df(v)$ points outside the sphere, we can say that $x_0$ can't be a maximum in $g$? – Let DC Jun 7 '15 at 19:27

Suppose $p \in U$ were a local maximum of $g = f_1^2 + f_2^2 + f_3^2$; then

$\nabla g(p) = 0; \tag{1}$

for any $x \in U$ we have

$\nabla g(x) = 2(f_1(x) \nabla f_1(x) + f_2(x) \nabla f_2(x) + f_3(x) \nabla f_3(x)); \tag{2}$

thus

$f_1(p) \nabla f_1(p) + f_2(p) \nabla f_2(p) + f_3(p) \nabla f_3(p) = \dfrac{1}{2} \nabla g(p) = 0. \tag{3}$

We may write the derivative $Df = D(f_1, f_2, f_3)$ in terms of the $\nabla f_i$, $1 \le i \le 3$, as the matrix

$Df = (\nabla f_1, \nabla f_2, \nabla f_3). \tag{4}$

The hypothesis that $f$ is of rank $3$ at every point is essentially the assertion that the rank of $Df$ is also $3$ everywhere; thus the vectors$\nabla f_i(x)$ must be linearly independent at every point $x \in U$. Granting this linear independence, (3) implies

$f_1(p) = f_2(p) = f_3(p) = 0, \tag{5}$

whence

$g(p) = 0; \tag{6}$

since $p$ is a maximum point of $g(x)$, we have for all $x \in U$

$0 \le f_1^2(x) + f_2^2(x) + f_3^2(x) \le g(p) = 0. \tag{7}$

But (7) is equivalent to

$f(x) = (f_1(x), f_2(x), f_3(x)) = 0 \tag{8}$

everywhere; that is, $f(x)$ is identically null, in contradiction to our hypothesis. Thus $g(x)$ may have no maxima in $U$. QED.

• Actually, the hypothesis on $f$ to be 'not identically null' is redundant since the OP assume $f$ to be has rank 3 at any point hence $f$ can not be a constant function. – Holonomia Jun 8 '15 at 19:16
• @Holonomia: true enough, that statement could have been extracted from the rank $3$ hypothesis; looks like our OP DC did it for us, and I took it as a gift! Cheers! – Robert Lewis Jun 8 '15 at 19:32
• Isn't $g = \sum_1^3 f_i^2$? Since the gradient operator is linear, $\nabla (u + v) = \nabla u + \nabla v$ for any functions $u$ and $v$; it follows from this. Recall each $f_i$ is a function from $U \to \Bbb R$. Does this clarify? If not, let me know. Cheers! – Robert Lewis Jun 11 '15 at 23:25
• Yes, I had even deleted the question because I remembered that the grad is linear, but thank you so much for your attencion. – Let DC Jun 11 '15 at 23:31
• @LetDC: glad to help out! – Robert Lewis Jun 11 '15 at 23:39

Sketch: Suppose to reach a contradiction that $g$ has a maximum at $a\in U.$ Then $\nabla g (a)$ is the zero vector. Show that this implies the nonzero vector $f(a)$ is perpendicular to the vectors $\partial f/\partial x_k, k=1,\dots , n.$ But $Df(a)$ has rank $3,$ hence these vectors span $\mathbb {R}^3.$ This is a contradiction.

• Where do you use the assumption that $U$ is connected? Cheers! – Robert Lewis Jun 7 '15 at 19:36
• You're right, it's not needed, thanks. – zhw. Jun 7 '15 at 19:38
• @zhw. Thanks... Cheers! – Let DC Jun 7 '15 at 19:42