Examples of bounded linear operators with range not closed

Define $T:L^1(\Bbb R)\to L^1(\Bbb R)$ by $Tf = gf$, where $$g(t)=\frac{1}{1+t^2}.$$Functions with compact support are dense in $L^1$, hence $T(L^1)$ is dense in $L^1$. But you can easily find an example showing that $T(L^1)\ne L^1$; hence $T(L^1)$ is not closed in $L^1$.

Take an operator $T\colon \ell_\infty \to c_0$ given by

$$T(\xi_n)_{n=1}^\infty = (\frac{\xi_n}{n})_{n=1}^\infty\quad (\xi_n)_{n=1}^\infty\in \ell_\infty).$$

Certainly $T$ is injective and has dense range. However, it cannot have closed range as it would be an isomorphism which is impossible as $\ell_\infty$ is non-separable.

This operator is actually compact. Compactness is a handy condition thay prevents operators whose range is not finite-dimensional to have closed range.

Take the bounded linear operator $$T: l^\infty \ni (x_j) \mapsto \left(\dfrac{x_j}{j}\right)\in l^\infty.$$

Then $x^{(n)}=(\sqrt{1},\sqrt{2},\dots,\sqrt{n},0,0,\dots)$ is in $l^\infty$ with $Tx^{(n)}=(1/\sqrt{1},\dots,1/\sqrt{n},0,\dots)$.

We have $Tx^{(n)} \to y=(1/\sqrt{j})$ in $l^\infty$, but $y \notin T(l^\infty)$ as $(\sqrt{j})\notin l^\infty$. So $T(l^\infty)$ is not closed.