Eigenvectors spanning closed subspace in a Banach space I'm looking for an example of a (bounded) linear operator $T$ on a Banach space $X$ with infinitely many eigenvalues such that $\sum_{\lambda\in\mathbb C}\ker(T-\lambda)$ is closed where the sum denotes the algebraic sum as vector spaces, i.e. the smallest vector space containing $\ker(T-\lambda)$ for all $\lambda\in\mathbb C$.
So far in the examples I've tried, it seems like this fails to be the case, at best this sum is dense in $X$, for instance any normal operators on a Hilbert space.
Since $\sum_{\lambda\in\mathbb C}\ker(T-\lambda)$ is closed, we know that it must have an uncountable dimension, so either there are uncountably many eigenvalues or the eigenspace has uncountable dimension. However this doesn't seem to help much.
My intuition is that $X$ being complete would prevent this from happening but I can't seem to show that there must be an element in the closure that is not an eigenvector.
So far my current attempt to show that $X$ need to have uncountably many eigenvalues is as follows:
Let $\lambda_i$ be the set of eigenvalues of $T$, i.e. $\{\lambda\in\mathbb C:\ker(T-\lambda)\neq\{0\}\}=\{\lambda_i\}_{i=0}^\infty$. Then we want to show that
$$\bigoplus_{i=0}^\infty\ker(T-\lambda_i)$$
is not closed.
Choose arbitrary $x_i\in\ker(T-\lambda_i)$ such that $\left\lvert x_i\right\rvert<i^{-2}$, then
$$\sum_{i=0}^\infty x_i\in\overline{\bigoplus_{i=0}^\infty\ker(T-\lambda_i)}$$
However
$$\alpha\sum_{i=0}^\infty x_i=\sum_{i=0}^\infty\lambda_ix_i$$
From here I'm hoping to show that $\sum_{i=0}^\infty x_i$ cannot be a finite sum of eigenvectors, but I'm not sure if the $\lambda_i$ on the RHS is the only way of expressing the LHS as a limit of scalar multiples of $x_i$.
 A: 
I'm looking for an example of a (bounded) linear operator T on a Banach space X with infinitely many eigenvalues such that $∑_{λ∈\mathbb C}\ker(T−λ)$ is closed where the sum denotes the algebraic sum as vector spaces, i.e. the smallest vector space containing $\ker(T−λ)$ for all $λ∈\mathbb C$.

Such operator does not exist. If $T$ has infinitely many eigenvalues, the subspace $∑_{λ∈\mathbb C}\ker(T−λ)$, as you defined it, is never closed. Indeed, suppose that $\{x_n\}$ is a sequence of unit eigevectors for $T$ with distinct eigenvalues $\{\lambda_n\}$. Let
$$\tag1
x=\sum_{n=1}^\infty 2^{-n}x_n. 
$$
Then $x\not\in\operatorname{span}\{x_n\}$, which would show that $\operatorname{span}\{x_n\}$ is not closed. Indeed, if we had $x\in \operatorname{span}\{x_n\}$, then there would exist eigenvectors $y_{1},\ldots,y_{m}$ and coefficients $\alpha_1,\ldots,\alpha_m\in\mathbb C$ with
$$\tag2
x=\sum_{k=1}^m\alpha_k y_{k}. 
$$
Comparing $(1)$ and $(2)$ lets incorporate the $y_1,\ldots,y_m$ to the sequence $\{x_n\}$, and hence we can  write
$$\tag3
\sum_{n=1}^\infty\,2^{-n}\,\beta_n\,x_n=0,
$$
where $\beta_n=1$ for all $n\geq\max\{n_1,\ldots,n_m\}$. We may now have repeated eigenvalues, but only finitely many of them. Without loss of generality, we may assume that $\beta_n\ne0$ for all $n$. Applying $T$ repeateadly to $(3)$ (and $T$ commutes with the limit because it is bounded) we obtain
$$\tag4
\sum_{n=1}^\infty\,2^{-n}\,\beta_n\,\lambda_n^k\,x_n=0,\qquad\qquad k\in\mathbb N. 
$$
Forming linear combinations of $(4)$ we get
$$\tag5
\sum_{n=1}^\infty\,2^{-n}\,\beta_n\,p(\lambda_n)\,x_n=0, \qquad\qquad p\in\mathbb C[x].
$$
Because the numbers $\beta_n$ and the numbers $\|x_n\|$ are uniformly bounded, the series in $(5)$ converge uniformly. Hence by taking limits we get
$$\tag6
\sum_{n=1}^\infty\,2^{-n}\,\beta_n\,f(\lambda_n)\,x_n=0, \qquad\qquad f\in C(\overline{\mathbb D}).
$$
Now consider a sequence  $\{f_n\}$ of continuous functions that converges pointwise to $f$ and such that  $|f_k|\leq c$ for all $k$. $\def\e{\varepsilon}$Fix $\e>0$. Then, using $f=1$ in $(6)$, there exists $n_0$ such that
$$
\Bigg\|\sum_{n>n_0}2^{-n}\beta_n x_n\Bigg\|<\frac\e{4c}.
$$
For $n=1,\ldots,n_0$, there exists $k_0$ such that $|f_k(\lambda_n)-f(\lambda_n)|<\frac\e{2\beta}$ for all $k\geq k_0$, where $\beta=\max\{|\beta_n|:\ n=1,\ldots,n_0\}$. Then, using $(6)$ and noting that, since $|f|\leq c$, the series with $f$ converges,
\begin{align}
\Bigg\|\sum_n2^{-n}\beta_n f(\lambda_n)x_n\Bigg\|
&=\Bigg\|\sum_n2^{-n}\beta_n f_k(\lambda_n)x_n-\sum_n2^{-n}\beta_n f(\lambda_n)x_n\Bigg\|\\[0.3cm]
&=\Bigg\|\sum_n2^{-n}\beta_n \big[f_k(\lambda_n)- f(\lambda_n)\big]\,x_n\Bigg\|\\[0.3cm]
&\leq\Bigg\|\sum_{n=1}^{n_0}2^{-n}\beta_n \big[f_k(\lambda_n)- f(\lambda_n)\big]\,x_n\Bigg\|+\frac\e2\\[0.3cm]
&\leq \sum_{n=1}^{n_0}2^{-n}|\beta_n|\, \big|f_k(\lambda_n)- f(\lambda_n)\big|\,\|x_n\|+\frac\e2\\[0.3cm]
&\leq  \big|f_k(\lambda_n)- f(\lambda_n)\big|\,\max\{|\beta_n:\ n=1,\ldots,n_0\}+\frac\e2\\[0.3cm]
&\leq\frac\e2+\frac\e2=\e.
\end{align}
As this can be done for any $\e>0$, we have shown that
$$\tag7
\sum_{n=1}^\infty\,2^{-n}\,\beta_n\,f(\lambda_n)\,x_n=0, \qquad\qquad f\in \overline{C(\overline{\mathbb D})}^p,
$$
where the $p$-closure means pointwise limits.
For any $m$, we can construct a sequence $\{f_k\}\subset C(\overline{\mathbb D})$ such that $f_k(\lambda_m)\to1$ and $f_k(\lambda)\to0$ for all $\lambda\ne\lambda_m$.  This would give us, from $(7)$
$$
2^{-m}\,\beta_m\,x_m=0, 
$$
a contradiction. In the few cases where we may have repeated eigenvalues, we get instead something of the form
$$
\sum_{j=1}^h2^{-m_j}\,\beta_{m_j}\,x_{m_j}=0,
$$
still a contradiction since the $x_{m_j}$ are linearly independent.
The contradiction shows that
$$
\sum_{n=1}^\infty 2^{-n}x_n\in \overline{\operatorname{span}}\{x_n\}\setminus \operatorname{span}\{x_n\}.
$$
