# [SOLVED]Proof that f is constant if it satisfies some relations

#### ianchenmu

##### Member
Let $V\subset \mathbb{R}^n$ be a connected, open set and $f:V\rightarrow \mathbb{R}^m$ be a map. Suppose that for all $x$,$y\in V$ ,
$|f(x)-f(y)|_*\leqslant|x-y|_\diamond ^\frac{3}2{}$,
where $| \ |_*$ (respectively $| \ |_\diamond$) is a norm on $\mathbb{R}^n$ (respectively $\mathbb{R}^m$). Prove that $f$ is constant.

(How could $f$ be constant? How to prove this? Thank you.)

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#### jakncoke

##### Active member
Let $V\subset \mathbb{R}^n$ be a connected, open set and $f:V\subset \mathbb{R}^m$ be a map. Suppose that for all $x$,$y\in V$ ,
$|f(x)-f(y)|_*\leqslant|x-y|_\diamond ^\frac{3}2{}$,
where $| \ |_*$ (respectively $| \ |_\diamond$) is a norm on $\mathbb{R}^n$ (respectively $\mathbb{R}^m$). Prove that $f$ is constant.

(How could $f$ be constant? How to prove this? Thank you.)
(Note: I'm just using some norm || ||, since all norms on finite dim vector spaces are equivalent)

well, $lim_{||h|| \to 0} \frac{||f(x+h) - f(x) - 0h||}{||h||} \leq ||h||^{\frac{1}{2}}$
(0 is the zero map from $V \to R^m$
cant we say as $||h|| \to 0$.$lim_{||h|| \to 0} \frac{||f(x+h) - f(x)||}{||h||} \to 0$.
So the linear 0 map from $$V \to \mathbb{R}^m$$ is the derivative of f(x) over this domain V. and since the linear map is 0, f(x) must be constant on V.

Am i missing something? I feel uneasy about the statement saying V is connected, i didn't use it anywhere here.

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Am i missing something? I feel uneasy about the statement saying V is connected, i didn't use it anywhere here.
You proved that f is constant on each disconnected part of V.
On disconnected parts those constants could be different.
However, since V is connected, the constant has to be the same everywhere.

#### jakncoke

##### Active member
I got it, a zero derivative need not imply a constant function unless. A zero derivative however does imply a constant function when the domain is connected.

Here, in connected open sets in $$\mathbb{R}^n$$ are path connected.
so. Continuing the above.

Since V is open and connected, it is path connected.

so for any x,y $\in V (1-t)x + ty$ for $t \in [0,1]$ is a path from x to y.

so f((1-t)x + ty)) = g(t)

so $\frac{||g(1) - g(0)||}{1} = 0$, or $||f(x)-f(y)|| = 0$, which implies that the components of f(x) and f(y) are equal, or, f(x) = f(y). Since x,y were arbitrary, f is constant over V.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
so for any x,y $\in V (1-t)x + ty$ for $t \in [0,1]$ is a path from x to y.

so f((1-t)x + ty)) = g(t)
Careful here.
It is not given that any x and y are connected through a straight path.
But then, you don't need that.

#### jakncoke

##### Active member
Careful here.
It is not given that any x and y are connected through a straight path.
But then, you don't need that.
yes you are right, path connectedness guarantees, some path $g:[0,1] \to V$, from x to y, such that g(1)=y, g(0)=x, which is all that is required to complete the above. thanks for the correction!.

#### ianchenmu

##### Member
You proved that f is constant on each disconnected part of V.
On disconnected parts those constants could be different.
However, since V is connected, the constant has to be the same everywhere.
I don't understand why jakncoke proved f is constant on each disconnected part of V ?
since the domain of f is V.

#### jakncoke

##### Active member
I don't understand why jakncoke proved f is constant on each disconnected part of V ?
since the domain of f is V.
Basically in my first post, i gave a proof that says that, since i let, $||h|| \to 0$, f(x) has 0 derivative in some neighborhood on x.

Now if i picked another point, y in our domain, again we have a 0 derivative in some neighborhood of y.

These two neighborhoods might be disjoint and thus disconnected.

So in the first neighborhood f(x) = $c_1$, and in the second neighborhood f(y) = $c_2$ but $c_1$ might not be equal to $c_2$.

For me to prove $f$ is constant on $V$, i need to prove $f(x) = f(y)$ for any two points x,y in V.

This is why, i needed that extra step and say "since V is connected it is path connected" ... This gives me a continous function from x to y. Since the function is continuous, (and since f is continous), it cannot break, meaning whatever constant value it takes on in the neighborhood of x, it must also take in the neighborhood of y, otherwise it wouldn't be continuous. Since x,y are arbitrary, f must be the same constant everywhere on V.

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