Let $X,Y$ be real Banach spaces, and let $B(X,Y)$ be the space of bounded linear operators.
Given $T \in B(X,Y)$ the modulus of $T$ is defined to be
$$
\gamma(T):=\inf \{ \,\|Tx\| \, \, | \, \, d(x,\ker T)=1 \}.
$$
It is known** that if the image of $T$ is closed in $Y$, then $\gamma(T)>0$.
I would like to find a counter-example for that, when we drop the completeness assumption from either of the spaces $X,Y$. That is, I want to build an example for two normed spaces $X,Y$ and a bounded operator $T:X \to Y$ with closed image, such that $\gamma(T)=0$. Of course, at least one of $X,Y$ should be incomplete.
Edit:
daw found a nice example when both $X,Y$ are non-complete. I wonder if there are examples when only one of the spaces is incomplete.
For intuition, when $X=Y=\mathbb{R}^n$ are endowed with the standard Euclidean norm, then for a real $n \times n$ matrix, $\gamma(A)=\min \{\sigma_i \, | \, \sigma_i \neq 0\} $ is the minimal non-zero singular value of $A$. (Of course, we should be careful with this intuition when working with infinite-dimensional spaces.)
A similar reasoning immediately shows that the condition on closed image is necessary, even when assuming both spaces are complete:
Indeed, take $X=Y=H$ to be an infinite dimensional separable Hilbert space, with orthonormal basis $e_i$, and set $Te_i=\frac{1}{i}e_i$. Then $T$ is injective and bounded, but for $e_n$, $|Te_n|=\frac{1}{n} \to 0$ when $n \to \infty$.
Thus,
$$
\gamma(T)=\inf \{ \,\|Tx\| \, \, | \, \, \|x\|=1 \}=0.
$$
Indeed, thinking in terms of "singular values", $\frac{1}{n}$ is a singular value of $T$, for every natural $n$.
Finally, the image of $T$ is easily seen to be non-closed:
Define $x_n=\sum_{i=1}^n e_i$. Then $Tx_n=(1,\frac{1}{2},\dots,\frac{1}{n},0,0,0,\dots) \in \text{Image}(T)$ converges to $(1,\frac{1}{2},\frac{1}{3},\dots) \notin \text{Image}(T)$. (It is "supposed to be" the image of $(1,1,1,\dots)$ which does not live in $H$).
**See theorem IV.1.6 in "Goldberg, Seymour. Unbounded linear operators: Theory and applications. Courier Corporation, 2006".
Best Answer
Take $X=Y=c_{00}$ (i.e., the space of sequences with finitely many non-zero elements) supplied with the $l^2$-norm. Define $$ Tx = (x_1, x_2/2,x_3/3,\dots,x_n/n,\dots). $$ Then $T$ is bijective, has unbounded inverse, and $\gamma(T)=0$.