Let $T:\mathbb{R}^{p} \rightarrow \mathbb{R}$ linear and $K=\left \{ \overline{x}\in \mathbb{R}^{p}:||x||_{2}\leq1 \right \}$. Show that T(K)=[-M,M] Let $T:\mathbb{R}^{p} \rightarrow \mathbb{R}$ linear (i.e. $T(x)=a_{1}x_{1}+\dots +a_{p}x_{p} $ )  and $K=\left \{ \overline{x}\in \mathbb{R}^{p}:||x||_{2}\leq1 \right \}$.
Show that $T(K)=[-M,M]$ with $M=\sqrt{(T(\overline{e}_{1}))^{2}+ \cdots + (T(\overline{e}_{p}))^{2}}$ where $\mathfrak{B}= \left \{ \overline{e}_{1}, ... , \overline{e}_{p} \right \}$ is the canonical basis. 
The linearity of $T$, symmetry of $K$ (i.e. $-K=K$), continuity of $T$, compactness of $K$ are useful.
What I've done so far:
I know that any  $\overline{x} \in \mathbb{R}^{p}$ can be written as 
$\overline{x}=b_{1}\overline{e}_{1}+ \cdots + b_{p}\overline{e}_{p}$
$\Rightarrow T(\overline{x})=T(b_{1}\overline{e}_{1})+ \cdots + T(b_{p}\overline{e}_{p})=b_{1}T(\overline{e}_{1})+ \cdots + b_{p}T(\overline{e}_{p})=b_{1}a_{1}+ \cdots +b_{p}a_{p}$      (*)
on the other hand 
$M=\sqrt{(T(a_{1}b_{1})^{2}+ \cdots + (T(a_{p}b_{p})^{2}}$                  (**)
I have the feeling that I can connect this (* and **) two using the Hölder inequality and then use symmetry to get $-M$ from $M$. is this a correct idea?

To find $M$:
$Max$ $T(x)$ 
    s.t. $||x||_{2}=1$ 
$\mathcal{L}=a_{1}x_{1}+ \cdots + a_{p}x_{p}-\lambda[(x_{1}^{2}+ \cdots + x_{p}^{2} )^{\frac{1}{2}}-1]$
F.o.c.
$x_{i}:\text{   } \frac{a_{i}}{\lambda}$ 
$\lambda: \text{   }(x_{1}^{2}+ \cdots + x_{p}^{2} )^{\frac{1}{2}}=1$
$\therefore x^{*}=T(\overline{e}_{1}+ \cdots + \overline{e}_{p})=T(\overline{e}_{1})+ \cdots + T(\overline{e}_{p})  $
$\therefore T(x^{*})=(T(\overline{e}_{1}))^{2}+ \cdots + (T(\overline{e}_{p})^{2})  $
Then $(T(x^{*}))^{\frac{1}{2}}=(\sum_{i=1}^{p}x_{i}^{*2})^{\frac{1}{2}}=||x||_{2}=1 \Rightarrow T(x^{*}))=\sum_{i=1}^{p}x_{i}^{*2}=1$
$\Rightarrow \text{   }M=T(x^{*})=(T(x^{*}))^{\frac{1}{2}}=\sqrt{T(\overline{e}_{1}))^{2}+ \cdots + (T(\overline{e}_{p})^{2}}$
and since for every function $min f=max -f$, We have that
$T(K)=[-M,M]$.
