Let $\rho(T)$ be the spectral radius of some compact bounded linear operator $T$ on a Banach space $(X,\|\cdot\|)$. Let $\|T\|_{op}$ be its operator norm. Then the famous Gelfand's formula says that $$\rho(T)=\lim_{k\to\infty}\|T^k\|_{op}^{1/k}=\inf_{k\in\mathbb N}\|T^k\|_{op}^{1/k}.$$ Here $$\|T^k\|_{op}=\inf\{c\geq0:\|T^kx\|\leq c\|x\|\ \forall x\in X\},$$ hence given $\epsilon>0$, we can find some $x_k\in X$ with $\|x_k\|=1$ such that $\|T^kx_k\|\geq\|T^k\|_{op}-\epsilon$.
I'm wondering whether (and if yes, under what conditions) it is possible to find an uniform element of $X$ satisfying this property. In other words, can we find some $x\in X$ with $\|x\|=1$ such that for all $k\in\mathbb N$, $\|T^kx\|\geq\|T^k\|_{op}-\epsilon$?
Heuristically, I'd like to apply the theorem of Riesz stating that we have countably many elements in $\sigma(T)$, having zero limit, with all nonzero elements being eigenvalues. Hence the spectral radius is attained at some eigenvalue; when working with a self-adjoint operator $T$, we may argue that this eigenvalue is real. Then we find an eigenvector $x$ corresponding to such a `large' eigenvalue, and hopefully this is such that $\|T^kx\|^{1/k}$ is close to $\|T^k\|_{op}$ for large $k$. If this cannot be done in general, maybe something can be proven along a subsequence of $(T^k)_{k\in\mathbb N}$? EDIT: It seems that after a few edits I have answered my own question, under the assumption of compactness and self-adjointness.
Any help is much appreciated.
The claim is not true.
Consider the Hilbert space $X=\mathbb{C}^3$ and the operator $$T(x_1,x_2,x_3)=(x_2,rx_3,0), 0<r<1$$ Then $$T^2(x_1,x_2,x_3)=(rx_3,0,0)$$ We have $\|T\|=1$ and $\|T^2\|=r.$ For $\|x\|=1$ we get $$\|Tx\|^2=|x_2|^2+r^2|x_3|^2\le |x_2|^2+r^2$$ Assume additionally that $\|Tx\|\ge 1-\varepsilon.$ Then $\|Tx\|^2\ge 1-2\varepsilon,$ and we get $$|x_2|^2\ge 1-2\varepsilon -r^2$$ Furthermore $\|T^2x\|=r|x_3|.$ Assume $\|T^2x\|\ge r-\varepsilon.$ Then $$|x_3|^2\ge \left (1-{\varepsilon\over r}\right )^2\ge 1 -{2\varepsilon\over r}$$ For $\varepsilon =r/4$ we have $$|x_2|^2\ge 1-{r\over 2}-r^2,\quad |x_3|^2\ge {1\over 2}$$ Taking $r={1\over 3}$ results in $$|x_2|^2={13\over 18}>{1\over 2},\quad |x_3|^2\ge {1\over 2}$$ which leads to a contradiction, as $\|x\|^2=1.$