Bounded inverse theorem

Remarks

The theorem follows from the open mapping theorem: a surjective continuous linear map between Banach spaces sends open sets to open sets. That theorem is proved with the Baire category theorem applied to the cover $Y = \bigcup_k \overline{T(B(0, k))}$. For a bijection, openness of $T$ is exactly continuity of $T^{-1}$.

A common application: if a vector space is complete for two norms $\norm{\cdot}_1$ and $\norm{\cdot}_2$ with $\norm{x}_2 \le c \norm{x}_1$, then the two norms are equivalent. Indeed, the identity map is a continuous bijection from the first norm to the second. Completeness of $Y$ is necessary. On $\ell^1$ the identity map from $(\ell^1, \norm{\cdot}_1)$ to $(\ell^1, \norm{\cdot}_\infty)$ is a continuous linear bijection, because $\norm{x}_\infty \le \norm{x}_1$. Its inverse is not bounded: $x = e_1 + \cdots + e_n$ has $\norm{x}_\infty = 1$ and $\norm{x}_1 = n$. Here $(\ell^1, \norm{\cdot}_\infty)$ is not complete.

Going deeper

For a student

The theorem is used to prove stability of solutions: if an equation $Tx = y$ has exactly one solution for each $y$, the solution depends continuously on $y$. It is equivalent, by short arguments, to the open mapping theorem and to the closed graph theorem. Together with the uniform boundedness principle these results all rest on the Baire category theorem.