Uniform unboundedness of linear operators

Question

Question: Suppose that $(T_k)_{k=1}^{\infty}$ is a sequence of invertible linear operators on $\mathbb{R}^n$. Suppose that $\forall x \in \mathbb{R}^{n}\setminus \{0\}$, we have $$\lim_{k\to\infty} \|T_k x\| = \infty .$$

Let $\mathbb{S}^{n-1}$ denote the unit sphere. Then does it follow that $$\lim_{k\to \infty} \inf_{x \in \mathbb{S}^{n-1}}\|T_k x\| = \infty? $$

Stated another way, is the sequence uniformly unbounded on the unit sphere?

Some of my thoughts so far:

(1) My question seems almost (but not quite) the converse of the Banach-Steinhaus theorem in the special setting of finite-dimensional vector spaces.

(2) Uniform unboundedness in the above question is equivalent to convergence of the sequence $$\lim_{k\to\infty} \sup_{x \in \mathbb{S}^{n-1}} \frac{1}{\|T_k x\|} = 0.$$ If I could prove that the sequence $(1/\|T_k x\|)_{n=1}^{\infty}$ (with $x$ in the unit sphere) was uniformly bounded and equicontinuous, then the Arzelà–Ascoli theorem together with pointwise convergence of $(1/\|T_k x\|)_{k=1}^{\infty}$ to zero would imply the result.

(3) Since we are working in $\mathbb{R}^n$, for each $k \in \mathbb{N}$ we may write $T_k$ in terms of its singular value decomposition (SVD): $T_k = U_k \Sigma_k V_k^T$. If $\|T_k x\|$ did not tend uniformly to $\infty$ on the unit sphere, then the smallest singular values of $\Sigma_k$ would have to be bounded, say, by $K > 0$. Denoting by $v_k^n$ the last column of $V_k$, we then have that $\forall k: \|T_k v_k^n\| < K$. Compactness of $\mathbb{S}^{n-1}$ would then allow me to extract a convergent subsequence from $(v_k^n)_{k=1}^\infty$, but I'm not sure how to make this useful.

(4) A geometric interpretation of my question is the following. For each $k$, the image of the unit sphere $T_k(\mathbb{S}^{n-1})$ is an ellipsoid $E_k$. The hypothesis that $\forall x \in \mathbb{R}^n\setminus \{0\}: \lim_{k\to \infty} \|T_k x\| = \infty$ implies that for any point $x \in \mathbb{S}^{n-1}$, the sequence of points $T_k x \in E_k$ grows unbounded in norm. It then seems geometrically intuitive to me that this can only happen if the length of the smallest "semi-axis" of the ellipsoids $E_k$ tend to $\infty$, which is equivalent to saying that $\lim_{k\to \infty} \inf_{x \in \mathbb{S}^{n-1}}\|T_k x\| = \infty. $ However, I haven't been able to turn this into a proof, and perhaps my intuition is misguided.

Why I care: I am reading the book Normally Hyperbolic Invariant Manifolds in Dynamical Systems by Stephen Wiggins. In deriving equation 3.5 while proving Lemma 3.1.1, he seems to be asserting the validity of the inference I am asking about.

Answer 1

Take $T_k x = k^2 x_1 -k x_2$. Then for any $x \neq 0$ we see that $|T_kx| \to \infty$.

However, $T_k ({1 \over \sqrt{1 + k^2}}(1,k)) = 0$ for all $k$.

Answer 2

Consider the case $n=2$, and take $$T_k = \pmatrix{k^2 \sin(1/k) & k^2 \cos(1/k)\cr 0 & 1\cr}$$ Write $x \in \mathbb S^1$ as $\pmatrix{\cos(\theta)\cr \sin(\theta)}$, and note that $(T_k x)_1 = k^2 \sin(\theta + 1/k)$. If $\sin(\theta) = 0$ this is $\pm k + O(1)$, while if $\sin(\theta) \ne 0$ it is $k^2 \sin(\theta) + O(k)$. So $\|T_k x\| \to \infty$ for any $x$. On the other hand, $$T_k \pmatrix{\cos(-1/k)\cr \sin(-1/k)} = \pmatrix{0 \cr \sin(-1/k)}$$ are bounded.