prove $T_a$ is a linear continuous function on X

Question

I'm wondering about this question. We have $n \geq 1$ and $X=\mathbb{R^n}$ with the 2-norm (euclidian norm). For $a=(a_k)_{1 \leq k \leq n}$ we have

$T_a : R^n \rightarrow R$

$x \rightarrow T_a(x)=\sum_{k=1}^{n}a_kx_k$ if $x=(x_k)_{1 \leq k \leq n}$

I would like to prove that $T_a$ is a linear continuous function on X and then deduce its norm, what is the way to proceed ?

Thank you very much

Answer 1

$T_a(x)$ is a dot $(x,a)$ of the vector $x$ with the vector $a$, so it is linear. For each vectors $x$ and $y$ of $\Bbb R^n$ we have Since $$|(x,a)-(y,a)|= |(x-y,a)|\le\|x-y\|\cdot \|a\|,$$

Thus if $a\ne 0$ then for each $\varepsilon>0$ if $\|x-y\|<\varepsilon/\|a\| $ then $|(x,a)-(y,a)|< \varepsilon$, so the map $T_a$ is continuous. If $a=0$ then the map $T_a$ is constant and therefore continuous too.

At last $$\|T_a\|=\sup\{T_a(x):\|x\|=1\}=\sup\{(x,a):\|x\|=1\} \le\sup\{\|x\|\cdot \|a\|:\|x\|=1\}= \|a\|.$$ From the other side, for $x=a/\|a\|$ we have $\|x\|=1$ and $T_a(x)=(a,a)/\|a\|=\|a\|$. Hence $\|T_a\|=\|a\|$.