Example of bounded below opeartor which is unbounded on Banach space?
In case of normed space we have $T:c_{00} \to \ell^{\infty}$ defined as $$T(x_{n})=(n x_{n})$$ is bounded below operator and unbounded. But we need unbounded operator defined on Banach space.
Inspired by this, let $\ell^1(\Bbb{R})$ be the Banach space of all $\Bbb{R}$-indexed sequences which are absolutely summable, equipped with the norm $$\|(x_\alpha)_{\alpha\in\Bbb{R}}\|_1 = \sum_{\alpha \in \Bbb{R}} |x_\alpha|.$$
The cardinality of $\ell^1(\Bbb{R})$ is $c$, this is because every $\Bbb{R}$-sequence in $\ell^1(\Bbb{R})$ is supported on a countable subset of $\Bbb{R}$. Since every vector space with dimension at least $c$ has dimension equal to its cardinality, it follows that $\dim \ell^1(\Bbb{R}) = c$. Therefore we can pick an $\Bbb{R}$-indexed algebraic basis $(v_\alpha)_{\alpha \in\Bbb{R}}$ for $\ell^1(\Bbb{R})$.
Define a linear map on this basis as $$T : \ell^1(\Bbb{R}) \to \ell^1(\Bbb{R}), \qquad T\left(\sum_{\alpha \in \Bbb{R}} \lambda_\alpha v_\alpha \right) = \left(\lambda_\alpha (\alpha^2+1) \|v_\alpha\|_1\right)_{\alpha \in \Bbb{R}}$$
We have \begin{align} \left\|T\left(\sum_{\alpha \in \Bbb{R}} \lambda_\alpha v_\alpha \right)\right\|_1 &= \left\|\left(\lambda_\alpha (\alpha^2+1) \|v_\alpha\|_1\right)_{\alpha \in \Bbb{R}}\right\|_1 \\ &= \sum_{\alpha \in \Bbb{R}} |\lambda_\alpha |(\alpha^2+1) \|v_\alpha\|_1 \\ &\ge \sum_{\alpha \in \Bbb{R}} |\lambda_\alpha |\|v_\alpha\|_1 \\ &\ge \left\|\sum_{\alpha \in \Bbb{R}} \lambda_\alpha v_\alpha\right\|_1 \end{align} so $T$ is bounded from below by $1$. On the other hand, we have $$\frac{\|T(v_\alpha)\|_1}{\|v_\alpha\|_1} = \alpha^2+1, \qquad \text{ for all }\alpha \in \Bbb{R}$$ so $T$ cannot be bounded.