Does a bounded from below linear operator with target set equal to the domain always have closed image? I'm using the definition of a bounded from below linear operator as a linear operator $X\overset{T}{{\to}}Y$, where $X$ and $Y$ are Banach spaces and the operator satisfies: for all $x \in X$, $\lVert T(x) \rVert \geq C \lVert x \rVert$, for some constant $C \gt 0$.
In this question i will refer only to such an operator in a Banach space $B$, $ T \colon B{{\to}}B$.
Note that I don't assume it to be continuous. With only these conditions, can we guarantee that the image of $T$ is closed in $B$? (Which is equivalent to asking if these assumptions imply continuity of the operator, by the open mapping theorem applied to the inverse of the operator, which exists since $\lVert T(x) \rVert = 0 \Rightarrow \lVert x \rVert = 0$).
I have tried to prove that it is closed with only these conditions and checked out other posts that have similar questions, but they don't seem to answer this one, at least in an immediate way. At this point, I believe that it is not generally true, but I also haven't been able to write a counter-example.
I appreciate any help and thank you in advance!
 A: Reardless of my comment above, the result is false, and it is in fact not so difficult to give a counter example.
Let $H$ be a separable, infinite dimensional Hilbert space with basis
$$
  E=\{e_n\}_{n\in \mathbb N}.
  $$
Consider  a Hamel basis for $H$ containing the linearly independent set $E$,  which we will write as the disjoint union
$$
  E\cup F.
  $$
Defining
$$
  N = \{(e_n, ne_n):n\in \mathbb N\},
  $$
we leave for the reader to verify  that the subset $C$ of $H\times H$ defined below is  linearly independent:
$$
   C = (0\times  E) \cup  (0\times  F) \cup  N.
  $$
Next consider a Hamel basis $ D$  of $H\times H$ containing $ C$, which we write as the disjoint union
$$
   D =\underbrace{(0\times  E) \cup  (0\times  F) \cup  N}_{ C} \ \cup  \ M,
  $$
and define
$$
  K= \text{span} (N\cup M).
  $$
where we insist that the span refers to the linear span,  no closure.
It is then easy to see that
$$
  H\times H = (0\times H) \oplus  K,
  $$
where "$\oplus $" is to be interpreted as algebraic direct sum, as opposed to orthogonal
direct sum.
For that reason, the canonical projection $\pi _1: H\times H\to H\times 0$,  once restricted to $K$ becomes a bijection,  and
hence we may speak of its inverse
$$
  \pi _1^{-1}: H\times 0 \to  K\subseteq H\times H.
  $$
Being the inverse of a bounded bijective map,  it is clear that   $\pi _1^{-1}$ is bounded below,  but it is certainly
discontinuous since
$$
  \|\pi _1^{-1}(e_n)\| = \|(e_n,  ne_n)\| = \sqrt{1+n^2}.
  $$
As noted by the OP, the range of   $\pi _1^{-1}$ cannot be closed.
