# Who can find an example of a bounded linear operator having a discontinuous inverse?

I need to find an example of a bounded linear operator having a discontinuous inverse.

I know that such operator exists. But I cannot find a concrete example.

## 4 Answers

Take a vector space $V$ with two topologies on it (satisfying reasonable additional axioms if you want them), one of which is a strict refinement of the other (for example, uniform convergence and $L^2$ convergence on $C([0, 1])$), and take the identity operator $V \to V$ where the source copy of $V$ has the finer topology and the target copy of $V$ has the coarser topology.

• This is in general an answer to a slightly different question, with “bounded” replaced by “continuous”; however I think the specific example suggested, of comparing the uniform and $L^2$ metrics, does aswer the question as written. Severin’s nice example answers a harder question, where the source and target are required to be the same normed vector space. – Qiaochu Yuan Jan 1 '18 at 23:14

Let $c_{00}=\{ (x_n)_{n\in \mathbb{N}} \subseteq \mathbb{R} \ : \ \text{only finitely many elements are nonzero} \}$. Then the operator

$$A: c_{00} \rightarrow c_{00}, (x_n)_{n\in \mathbb{N}} \mapsto \left(\frac{1}{n} x_n \right)_{n\in \mathbb{N}}$$

is a bijection and has operator norm 1. However, the inverse is unbounded.

• See other examples in (math.stackexchange.com/q/710142) – Jean Marie Jan 1 '18 at 23:35
• I upvoted this response. Maybe it's worth pointing out that, by analogy with Fourier series, if $X_{\epsilon} = \{f \in C^{\infty}([0,1]) \, \mid \, \forall t \in [0,\epsilon] \, \, f(t) = 0\}$ with the uniform norm, then $A : X \to X$ given by $(Af)(x) = \int_{0}^{x} f(s) \, ds$ works in a similar way. – fourierwho Jan 2 '18 at 2:59
• @fourierwho Why don't you make this an answer? I think it's a very nice example. – Severin Schraven Jan 2 '18 at 10:35

You should note that if $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear, continuous and invertible then it follows that $T^{-1}$ is continuous.

I'll try to give a "continuous" version of Severin Schraven's post. Those familiar with Fourier series will see the analogy.

Let $X$ be the vector space of trigonometric polynomials with positive frequencies, considered as a subspace of $C([0,1])$. In other words, $X = \text{span}\{e_{n} \, \mid \, n \in \mathbb{N}\}$, where $e_{n} : [0,1] \to \mathbb{C}$ is given by $$e_{n}(x) = e^{i 2 \pi n x}.$$

Next, define $A : X \to X$ by $$[A(f)](x) = \int_{0}^{x} f(s) \, ds - \int_{0}^{1} \int_{0}^{x} f(s) \, ds \, dx.$$ One can check that this is well-defined (i.e. maps into $X$). In fact, $A(e_{n}) = \frac{1}{i 2 \pi n} e_{n}$ so that this is just like Severin's example. This also shows that $A$ is invertible and has norm $1$.

There are many other similar examples. Consider $C^{\infty}_{c}((0,1])$, i.e. smooth functions on $(0,1]$ with compact support. Let $A : C^{\infty}_{c}((0,1]) \to C^{\infty}_{c}((0,1])$ be given by $$[A(f)](x) = \int_{0}^{x} f(s) \, ds.$$ Because of the compact support assumption, this is well-defined and invertible. The same example works if we replace $C^{\infty}_{c}((0,1])$ by $\{f \in C^{\infty}([0,1]) \, \mid \, \forall t \in [0,\epsilon] \, \, f(t) = 0\}$ for some $\epsilon \in (0,1)$.