# $f\colon\mathbb R^n \rightarrow \mathbb R$ be a linear map with $f(0,0,0,\ldots,0)=0.$

I was thinking about the following problem :

Let $f\colon\mathbb R^n \rightarrow \mathbb R$ be a linear map with $f(0,0,0,\ldots,0)=0.$ Then the set $\{f(x_1,x_2,\ldots,x_n):\sum_{j=1}^{n}x_j^2 \leq 1\}$ equals to which of the following:
$1.[-a,a]$ for some $a \in \mathbb R,a \geq 0.$
$2.[0,1]$.
$3.[0,a]$ for some $a \in \mathbb R,a \geq 0$
$4.[a,b]$ for some $a,b \in \mathbb R, 0<a \leq b .$

My Attempt: Since $f\colon\mathbb R^n \rightarrow \mathbb R$ is a linear map, $(a) f(u+v)=f(u)+f(v),$ for $u,v \in \mathbb R^n$ and $(b)$ for $\lambda \in \mathbb R, f(\lambda v)=\lambda f(v)$ .Now Using @David Mitra's suggestion, we see that if $c=(c_1,c_2,\ldots,c_n)$ is in the set then so is $-c.$ Now, I see $f\{c+(-c)\}=f\{(c_1,c_2,\ldots,c_n)+(-c_1,-c_2,\ldots,-c_n)\}=f\{(c_1-c_1,c_2-c_2,\ldots,c_n-c_n)\}=f(0,0,\ldots,0)=0$.

Also,since $f$ is linear,$f\{c+(-c)\}=f(c)+f(-c)$ and hence $f(c)+f(-c)=0$ and so $f(-c)=-f(c).$

So, if for some $c \in \mathbb R^n, f(c)=a$,then $f(-c)=-a.$ So option $(1)$ seems to be right .

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Hint: If $c$ is in the set, then so is $-c$. – David Mitra Feb 7 '13 at 3:14
A linear map always sends $0$ to $0$. – 1015 Feb 7 '13 at 3:58

Topological proof that 1 is correct.

Since linear maps are automatically continuous in finite dimension, $f$ is continuous.

The set you are looking for is the image of $B$, the closed unit ball centered at $0$ with respect to the Euclidean norm.

This set is compact and connected.

So $f(B)$ is a compact connected set of $\mathbb{R}$. Hence $f(B)=[a,b]$.

Now by David Mitra's comment, you see that $f(B)$ is symmetric with respect to $0$.

So $f(B)=[-a,a]$.

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Hint:

If $x=(x_1,\ldots,x_n)$ satisfies $\sum\limits_{i=1}^n x_i^2$, then so does $-x$. Using the linearity of $f$ allows you to immediately rule out three of the options.

To show that the remaining option is correct requires a bit more work...

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