The inverse of bounded operator? Is the inverse of a bounded operator always bounded , if yes how to prove it ?
If You consider an invertible, i.e. bijective and bounded linear operator $A:X\rightarrow Y$, between two Banach-spaces (it´s important they are complete), then as a consequence of Baires category theorem A is open (open mapping theorem) and so $A^{-1}:Y \rightarrow X$ is continuous, i.e. bounded.
It depends on how you define "inverse" of an operator. In general, if you don't require it be bounded, then no. For example, consider the bijection from $\mathbb{R}$ onto $(-1,1)$. But sometimes we say an operator is invertable only if we can find a bounded operator such that the composition is identity.