Exercise about an operator $\ell^2 \to \mathbb{R}$ Let $T_n : \ell^2 \to \mathbb{R}$ such that: $$T_n(x) = \sum_{k=1}^{n} \alpha_kx_k.$$

Prove that $T_n$ is continuous and find its norm.

My attempt: 
I have proved that $|T_n(x)|\leq M\|x\|_{2}$ such that $M=\left( \sum_{k=1}^n\alpha_k^2\right)^{1/2} $
so $T_n$ is  continuous and $\|T_n\|\leq M$. I have to prove $\|T_n\|\geq M$ now.
Can someone help me Please?
 A: Because you wrote $T_n : \ell^2 \to \mathbb{R}$ with $T_n(x) = \sum_{i=1}^n \alpha_i x_i,$ I'm assuming that $\ell^2$ is a real Hilbert space with the usual inner product (the general argument works if $\ell^2$ is complex, although you'll need to add complex conjugations where appropriate) and I'm assuming that $n$ is an integer (the argument below can be extended to the case where $n = \infty$ but you'll have to assume that $\left(\alpha_1, \alpha_2, \ldots\right) \in \ell^2$ and also justify series convergence where necessary).
Proof of continuity: For every $i \in \mathbb{N},$, let $e_i = (0, \ldots, 0, 1, 0, \ldots)$ be the standard orthonormal basis for $\ell^2$ (that is, $e_i$ is $1$ at its $i^{th}$ coordinate and $0$ everywhere else) and let $\pi_i : \ell^2 \to \mathbb{R}$ be $\pi_i(x) := x_i = \langle e_i, x \rangle$, which is a continuous linear functional (by the Cauchy-Schwarz inequality). Then for any $x \in \ell^2,$
$$T_n(x) = \sum_{i=1}^n \alpha_i x_i = \alpha_1 \pi_1(x) + \cdots + \alpha_n \pi_n(x),$$
which shows that $T_n = \alpha_1 \pi_1 + \cdots + \alpha_n \pi_n$ is a finite sum of (scalar multiples of) continuous linear functionals and so is itself a continuous linear functional.
Finding the norm: In fact, let $\alpha := a_1 e_1 + \cdots + a_n e_n,$ which is an element of $\ell^2$ with norm $\|\alpha\|_2 = \sqrt{\sum_{i=1}^n \alpha_i^2}$. Then
$$\langle \alpha, x \rangle = a_1 \langle e_1, x \rangle + \cdots + a_n \langle e_n, x \rangle = T_n(x)$$
shows that $T_n = \langle \alpha, \,\cdot\, \rangle$ (side note: this also proves that $T_n$ is continuous, although the proof above generalizes better to non-Hilbert spaces) so by the Riesz representation theorem, $\|T_n\|_2 = \|\langle \alpha, \,\cdot\, \rangle\|_2 = \|\alpha\|_2.$ Thus $\|T_n\|_2 = \sqrt{\sum_{i=1}^n \alpha_i^2}$.
