$I+T$ is not bounded below Let $\mathcal{H}$ a separable Hilbert space. I want to show that the operator $I+S \in L(\mathcal{H})$, with $S\in L(\mathcal{H})$ the left shift, is not bounded below; i.e, there exists a sequence $\{x_n\}_{n\in \mathbb{N}}\subset \mathcal{H}$ such that $||x_n||=1$ and $T(x_n)\underset{n \longrightarrow \infty}{\longrightarrow} 0$.
My aproach was:
I couldn't find the suitable sequence, then i tried the following...
$\sigma(S)=\mathbb{D} \implies S+I \notin \mathcal{G}_l(\mathcal{H})$. That's implies $I+S$ isn't bounded below or $I+S^*$ isn't bounded below. I tried to find a lower bound for $I+S^*$, but i couldn't.
Thanks for read.
 A: Assume $I+S$ is bounded below i.e.$$\|(I+S)x\|\ge c\|x\|$$ Then the range of $I+S$ is closed. As $-1\in \sigma(S)$  the range of $I+S$ cannot be equal $\mathcal{H}.$ Thus $\ker (I+S^*)={\rm Im}(S)^\perp\neq \{0\},$ a contradiction, as $-1$ is not an eigenvalue of $S^*.$ The same reasoning is valid for $I+S^*.$
If we are after a concrete sequence of vectors, let
$v_t(k)=t^k$ for $|t|<1.$ Then
$(I+S)v_t=(1+t)v_t.$ Hence $${\|(I+S)v_t\|\over \|v_t\|}=1+t\underset{t\to -1^+}{\longrightarrow}0$$
Concerning $I+S^*$ we have $$(I+S^*)v_t=(1,t+1,t^2+t,\ldots, t^n+t^{n-1},\ldots)$$
Thus $$\|(I+S^*)v_t\|^2=1+(1+t)^2\|v_t\|^2$$
Hence
$${\|(I+S^*)v_t\|^2\over \|v_t\|^2}={1\over \|v_t\|^2}+(1+t)^2 \underset{t\to -1^+}{\longrightarrow}0$$
A: For each $n\in\Bbb N$ define $x_n=(x_{n,k})_k\in\mathcal H=\ell^2(\Bbb N)$ by
$$x_{n,k}:=\begin{cases}\frac{(-1)^k}{\sqrt n}&\text{ if }1\le k\le n\\0&\text{ if }k>n.\end{cases}$$
Then, $\|x_n\|=1,$ and $T(x_n)\to0$ since
$$T(x_n)_k:=\begin{cases}\frac{(-1)^n}{\sqrt n}&\text{ if }k=n\\0&\text{ else.}\end{cases}$$
