# Set of surjective operator in normed space

Let $X$, $Y$ are Banach space, then set of surjective bounded linear operators is open in $B \left(X,Y \right)$.

I think that if X and Y are normed space (incomplete) then this is not true.

let $T$ = $id$

$$T: \left(C[-1,1],||.||_{\infty} \right) \rightarrow \left(C[-1,1], ||.||_1 \right)$$ T is bijection with unbounded inverse.

In one hand $T_n = \left(1- \dfrac{1}{n+1} \right) T$ is sequence of bijective operators with unbounded inverse too.

In other hand we can define $T_n^{'} \left( f \right) = f (x) - \dfrac{1}{n} f(-x)$. These operators are not surjective, but bound and linear. Then T is limit point.

Only I doubt so, if I change the norm in the second space and make it a Banach one, then nothing seems to change, but this statement will be wrong.

Thank you!

## 1 Answer

When you make the codomain Banach with the infinity norm, your $T_n'$ are invertible. You can write $$T_n'=I-\frac1n\,R,$$ where $Rf (x)=f (-x)$. You have $\|R\|=1$, so $R_n=\frac1n\,R$ satisfies $\|R_n\|<1$, and so the Neumann series $\sum_kR_n^k$ converges to $(I-R_n)^{-1}$.

So $T_n'$ is surjective regardless of the norm you choose in the codomain.

What you could do is to take $$g_n (x)=\begin {cases}n+n^3x,&\ x\in [-1/n^2,0]\\ \ \\n-n^3x,&\ x\in [0,1/n^2]\\ \ \\0,|x|>1/n\end {cases}$$ and put $S_nf=f-g_nf$. Then $$\|(I-S_n)f\|_1=\int_{-1}^1g_nf\leq2\|f\|_\infty/n\to0,$$so $S_n\to I$. And $S_n$ is not surjective because $$S_nf\left (\frac {1-n}{n^3}\right)=0.$$

• Glad I could help :) – Martin Argerami Feb 9 '18 at 16:14