Limits in a normed space

Question

Let $X$ be a normed space. Suppose that $x,y \in X$ and $x+y \neq 0$.

If $\lim_{n \to \infty}\|(x+y)^n\|^{\frac{1}{n}} = 0$, then is it true that $\lim_{n \to \infty} \|x^n\| ^{\frac{1}{n}} = 0$ ?

Answer 1

As noted by Nicolas, $x^n$ does not make sense in a general normed space.

For linear operator $T$ on a Banach space, $\lim_{n \rightarrow \infty} ||T^n|| ^{\frac{1}{n}}$ computes the "spectral radius". So we could ask, given two operators $S,T$, with $S-T \ne 0$, is it true that if the spectral radius of $S+T$ is zero, then must the spectral raius of $T$ be zero?

Answer, no.

For example, even with $2 \times 2$ matrices this fails. $$ S := \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} \\ T := \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \\ S+T := \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} $$ Then $S+T$ is not $O$, but has spectral radius $0$. Easy computation, since $(S+T)^n = O$ for $n \ge 2$.
But $T$ has spectral radius $1$. Also easy, since $T^4 = I$, the powers repeat with period $4$ and $\|T^n\| = 1$ for all $n$.

Answer 2

I'm assuming you're actually working in a normed algebra.

The spectral radius of $x\in X$ is defined as $$r(x) = \lim_{n\to\infty} \|x^n\|^{\frac1n}$$

It is a standard result that the spectral radius is radius of the smallest closed ball around the origin enclosing the spectrum of $x$, i.e. $$r(x) = \max\{|\lambda| : \lambda \in \sigma(x)\}$$

So you are interested in whether $x+y\ne 0$ and $r(x+y) = 0$ implies $r(x) = 0$.

The answer is no. For a counterexample consider the $2\times 2$ matrices $x = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ and $y = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$.

Clearly $\sigma(x) = \{1\}$ and $\sigma(y) = \{-1\}$, but $$\sigma(x+y) = \sigma\left(\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\right) = \{0\}$$

Therefore $r(x+y) = 0$ but $r(x) = 1$ and $r(y) = -1$.