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Let $f:\mathbb R^m \to \mathbb R^m$ be a differentiable function. let $Df(x)$ be the derivative of $f$ at $x\in \mathbb R^m$. which of the following is/are correct?

(a)$Df(0)(u)=0, \forall u \in \mathbb R^m$

(b)$Df(x)(u)=0, \forall u \in \mathbb R^m$ and some $x\in \mathbb R^m$ only if $f$ is a constant.

(c)$Df(x)(u)=0, \forall u \in \mathbb R^m$ and all $x\in \mathbb R^m$ only if $f$ is a constant.

(d)If $f$ is not a constant function, then $Df(x)$ is a one-one function for some $x\in \mathbb R^m$.

(a) Need not true. since at $0\in \mathbb R^m$, $Df(0)=O$(zero transformation). analogous to derivative in real variable. it may have non zero linear transformation as a derivative at $0$. But I couldn't find the counterexample.

(b) I am not able to find the counter example.

(c)True, analogous to the result $f'(x)=0$ in a connected domain, then $f(x)=c$. but I am not able to prove the generalized result. I know that $\mathbb R^m$ is connected.

(d) If $f$ is not a constant function, then $Df(x)$ is a linear transformation. How to prove that it is one-one? for finite dimensional transformation one-one and onto are equivalent. Please help me

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  • $\begingroup$ For (b), what about $y_i = \sum_1^m x_j^2$, $1 \le i \le m$; i.e $(f(x))_i = y_i$. $\endgroup$ Dec 14 '17 at 16:57
  • $\begingroup$ In (c), do you mean $Df(x)(u) = 0$? $\endgroup$ Dec 14 '17 at 17:01
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    $\begingroup$ yes. in b and c. I apoogize for the error. $\endgroup$
    – user464147
    Dec 14 '17 at 17:02
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    $\begingroup$ I will correct it. $\endgroup$
    – user464147
    Dec 14 '17 at 17:02
  • $\begingroup$ @RobertLewis sir, I don't understand the counterexample for (b). can you please help me. $\endgroup$
    – user464147
    Dec 14 '17 at 17:09
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a.) Take $f(x) = Ax$ for some $m \times m$ matrix $A \ne 0$; then $Df(x)(u) = Au$ for all $x \in \Bbb R^m$, i.e. $Df(x) = A \ne 0$. So $Df(0)(u) \ne 0$ for some $u$;

b.) Consider the function $f:\Bbb R^m \to \Bbb R^m$ defined by

$f(x) = \begin{pmatrix} \sum_1^m x_i^2 \\ \sum_1^m x_i^2 \\ \vdots \\ \sum_1^m x_i^2 \end{pmatrix}, \tag 1$

where

$x = (x_1, x_2, \ldots, x_m) \in \Bbb R^m; \tag 2$

that is, each component of $f(x)$ is the same function $\sum_1^m x_i^2$ of $x$. Then

$Df(x) = \begin{bmatrix} 2x_1 & 2x_2 & \ldots & 2x_m \\ 2x_1 & 2x_2 & \ldots & 2x_m \\ \vdots \\ 2x_1 & 2x_2 & \ldots & 2x_m \end{bmatrix}; \tag 3$

that is, every row of $Df(x)$ is the same vector $(2x_1, 2x_2, \ldots, 2x_m)$; then

$Df(0) = 0, \tag 4$

so

$Df(0)(u) = 0, \; \forall u \in \Bbb R^m, \tag 5$

but $f(x)$ is not constant;

c.) here we use integration; for $x \in \Bbb R^m$ consider the path

$a(t) = tx, \; 0 \le t \le 1; \tag 6$

then

$f(x) - f(0) = \displaystyle \int_0^1 \dfrac{df(a(s))}{ds}ds = \int_0^1 Df(a(s)) \left (\dfrac{da(s)}{ds} \right ) ds = \int_0^1 0 \; ds = 0, \tag 7$

since

$Df(a(s)) = 0; \tag 8$

thus, for all $x \in \Bbb R^m$,

$f(x) = f(0), \tag 9$

and $f(x)$ is constant.

d.) take

$f(x) = Ax, \tag{10}$

where $A \ne 0$ is singular $m \times m$ matrix. Then $f(x) \ne 0$ but

$Df(x) = A \tag{11}$

is nowhere one-to-one, since $\ker A \ne \{0\}$.

In short, only item (c) is true; (a), (b) and (d) are false.

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