Zero is an interior point of a surjective bounded operator with nonempty kernel

Question

Let $T$ be a linear bounded operator on a Banach space $X$ over $\mathbb{C}$, and $T$ is surjective but not injective. Then $0$ is an interior point of the spectrum of $T$.

I came across this question a few days ago. It has been closed, but I want to check my proof of the statement.

Proof

Preparations: $\ker T := Y$ is a closed subspace of $X$. We can take a factor $X/Y$ and the respective quotient map $Q$. It is well-known that $X/Y$ is Banach space w.r.t. the quotient norm $||x + Y|| = \inf_{y \in Y}||x + y||$. By the inverse mapping theorem $S := TQ$, which is bijective, has a bounded inverse. Denote $r = 2||S^{-1}|| > 0$. By the definition of the quotient norm $\forall x \in X$ $\exists x' \in X$ s.t. $Tx' = x$ and $||x'|| \leq r ||x||$. For any given $x \in X$ we will denote the corresponding $x'$ as $R(x)$. There are many vectors that fits, but we will assume that only one is chosen for each $x \in X$.

Main part:

Fix an arbitrary $\lambda \in \mathbb{C}$ s.t. $|\lambda| \le (3r)^{-1}$. Fix an arbitrary $x_0 \in Y$ s.t. $||x_0|| = 1$. Denote $x_n = \lambda^n R^n(x_0)$, $n \in \mathbb N$. By the definition of $R$ $||x_n|| \leq (\lambda r)^n \le 3^{-n}$.

Consider an element $w \in X$ defined by the following absolutely convergent series: $$ w = \sum_{n = 0}^\infty (-1)^n x_n. $$ Since $$ \left|\left|\sum_{n = 1}^\infty (-1)^n x_n\right|\right| \leq \frac 12, $$ $w$ is a nonzero vector.

Now we observe that $$ (T + \lambda E) w = \sum_{n = 0}^\infty (-1)^n T x_n + \sum_{n = 0}^\infty (-1)^n \lambda x_n = \sum_{n = 1}^\infty (-1)^n \lambda x_{n-1} + \sum_{n = 0}^\infty (-1)^n \lambda x_n = 0. $$

Hence $\ker (T + \lambda E) \neq 0$, so $-\lambda$ belongs to the spectrum of $T$ (moreover, it's in the point spectrum). Q.E.D.

Answer 1

Your argument is good. I'll just comment a few minor things.

• The map $S$ is not $TQ$. What you want to do is define $S(x+Y)=Tx$. This map is bounded and bijective, so invertible.

• You can take $r=\|S^{-1}\|$, you don't need the 2. You choose $Rx$ to be a representative of $S^{-1}x$ such that $\|Rx\|\leq\|S^{-1}x\|$. Then $$\|Rx\|\leq\|S^{-1}x\|\leq\|S^{-1}\|\,\|x\|=r\|x\|.$$

• You can take $|\lambda|< (2r)^{-1}$. Because this makes the sum starting at $k=1$, less than 1, while the first term has norm 1.

• You don't need the $(-1)^n$, they play no role.

• Your last computation will look nicer (to me, at least), if you write it in the form $Tw=\lambda w$ (after removing the $(-1)^n$).

Answer 2

In order to make the conclusion, I post an improved version of the proof.

Let $T$ be a linear bounded operator on a Banach space $X$ over $\mathbb{C}$, and $T$ is surjective but not injective. Then $0$ is an interior point of the spectrum of $T$.

Proof

Preparations: $\ker T := Y$ is a closed subspace of $X$. We can take a factor $X/Y$ and the respective quotient map $Q$. It is well-known that $X/Y$ is Banach space w.r.t. the quotient norm $||x + Y|| = \inf_{y \in Y}||x + y||$. By the universal property of quotients there exists (a unique) $S \in \mathscr B (X/Y, X)$ s.t. $SQ = T$. Since $T$ is surjective, $S$ is bijective, so by the inverse mapping theorem $S$ has a bounded inverse. Fix an arbitrary $\varepsilon > 0$ and denote $r_\varepsilon = (1+\varepsilon)||S^{-1}|| > 0$. By the definition of the quotient norm $\forall x \in X$ $\exists x' \in X$ s.t. $Tx' = x$ and $||x'|| \leq r_\varepsilon ||x||$. For any given $x \in X$ we will denote the corresponding $x'$ as $R_\varepsilon(x)$. There are many vectors that fits, but we will assume that only one is chosen for each $x \in X$.

Main part:

Fix an arbitrary $\lambda \in \mathbb{C}$ s.t. $|\lambda| < (2r_\varepsilon)^{-1}$. Fix an arbitrary $x_0 \in Y$ s.t. $||x_0|| = 1$. Denote $x_n = \lambda^n R^n_\varepsilon(x_0)$, $n \in \mathbb N$. By the definition of $R_\varepsilon$ $||x_n|| \leq (\lambda r_\varepsilon)^n < 2^{-n}$.

Consider an element $w \in X$ defined by the following absolutely convergent series: $$ w = \sum_{n = 0}^\infty x_n. $$ Since $\left|\left|\sum_{n = 1}^\infty x_n\right|\right| < 1$, $w \neq 0$.

Now we observe that $$ T x_n = T \lambda^n R^n_\varepsilon(x_0) = \lambda^n R^{n-1}_\varepsilon(x_0) = \lambda x_{n-1}. $$ Hence $$ T w = \sum_{n = 0}^\infty T x_n = \sum_{n = 1}^\infty \lambda x_{n-1} = \lambda\sum_{n = 0}^\infty x_{n} = \lambda w. $$

Therefore, $\ker (T - \lambda E) \neq 0$, so $\lambda$ belongs to the spectrum of $T$. Moreover, we have proven that $\sigma_p (T) \supset \{\lambda \in \mathbb C \mid |\lambda| < (2||S^{-1}||)^{-1}\}$. Q.E.D.