$T^{-1}$ is not continuous The set $\ell_1=\{(x_j)_{j=1}^\infty : x_j\in\mathbb{R}, \sum_{j=1}^\infty |x_j|<\infty\}$ is a vector space over $\mathbb{R}$ and $T:\ell_1\to\ell_1$ defined by $$T(x_1,x_2,x_3,\dots)=(x_1/1,x_2/2,x_3/3,\dots)$$ is linear, continuous and invertible over $Im(T)$.
My doubt: Why $T^{-1}:Im(T)\to\ell_1$ is not continuous?
Could you please help me? Thanks.
 A: A generalization is that if $T$ is a linear operator on a normed space $X$, and there is a sequence of unit vectors $(e_n)$ in $X$ such that $T(e_n)\to 0$, then $T$ cannot have a continuous inverse.  If $T$ is injective then at least a linear inverse exists on the range of $T$, but then $\dfrac{\|T^{-1}(T(e_n))\|}{\|T(e_n)\|}\to \infty$, showing that $T^{-1}$ is not continuous.
The next thing to having nontrivial kernel, $T(v)=0$ with $v\neq 0$, is having $T(v)$ get arbitrarily small while $v$ remains bounded away from $0$.  In the nontrivial kernel case, $T$ has $0$ as an eigenvalue, and it is also said that $0$ is in the "point spectrum" of $T$.  In the $T(v_n)\to 0$ while $v_n$ doesn't case, $T$ is said to have an "approximate eigenvalue" $0$, and $0$ is in the "approximate point spectrum" of $T$.
The property of an operator where this doesn't happen, allowing the inverse to be continuous, is called "bounded below".  $T$ is called bounded below if there exists $c>0$ such that for all $x\in X$, $\|Tx\|\geq c\|x\|$.  When that happens, the inverse $T^{-1}$ is bounded on the image of $T$ with norm at most $\frac1c$.  
