A Lemma Related to Oseledets' Theorem

Question

Let $(X,\mu)$ be a probability space, $T:X\to X$ measurable and preserves $\mu$, $A:X\to GL_d(\mathbb{R})$ measurable and $\log^+\Vert A\Vert$, $\log^+\Vert A^{-1}\Vert$ integrable. Define $A^1(x)=A(x)$,

$$A^n(x)=A(T^{n-1}x)A^{n-1}(x)=A(T^{n-1}x)\ldots A(x),\qquad n>1$$

and

$$\lambda(x,v)=\lim\limits_n\sup\frac{1}{n}\log\Vert A^n(x)v\Vert$$

The lemma (from Viana's Lectures on Lyapunov Exponents, Lemma 4.4.(iii), p.40) is:

Lemma: $\lambda(x, v+v')=\max\{\lambda(x, v),\lambda(x, v')\}$ if $v+v'\ne 0$.

One side is easy using the fact that limsup and max can be changed in series, thus \begin{align} \max\{\lambda(x, v),\lambda(x, v')\}&=\lim\limits_n\sup\frac{1}{n}\log(\Vert A^n(x)v\Vert+\Vert A^n(x)v'\Vert)\\ &\geq \lim\limits_n\sup\frac{1}{n}\log(\Vert A^n(x)v+ A^n(x)v'\Vert)=\lambda(x, v+v') \end{align} The other side, I thought that maybe $\Vert{Av}\Vert+\Vert{Av'}\Vert\leq c(\Vert{Av}+{Av'}\Vert )$ if $v+v\ne 0, A\in GL_d(R)$, however it's not true as in dim 2 we can use diagonal matrix to make their angle approach $\pi$.

Answer 1

The statement is false.

For a counterexample take say $A(x)=\begin{pmatrix}2 & 0\\0&3\end{pmatrix}$. For $v=(1,1)$ and $v'=(0,-1)$ we have $$ \lambda(x,v)=\lambda(x,v')=\log3\quad\text{but}\quad \lambda(x,v+v')=\log2. $$

Answer 2

In a set of notes by J. Bochi, page pp. 5, the author states a similar statement to that of the OP's but with a slightly different assumption:

\begin{align} \lambda(x;v+v')\leq\max\{\lambda(x;v),\lambda(x,v')\}\tag{0}\label{zero} \end{align} with equality if $\lambda(x;v)\neq\lambda(x;v')$.

This statement and many others there are left as exercise. Here is proof

The OP has already established that \begin{align} \lambda(x;u+v)\leq\max(\lambda(x;u),\lambda(x;v))\tag{1}\label{one} \end{align}

Notice that $\lambda(x;\alpha v)=\lambda(x;v)$ for all scalar $\alpha\neq0$ and vector $v\neq0$.

Suppose $\lambda(x;u)\neq\lambda(x;v)$ and w.l.g., $\lambda(x;v)<\lambda(x,u)$ so that $\max(\lambda(x;u),\lambda(x;v))=\lambda(x;u)$. Applying \eqref{one} yields $$\lambda(x;v+u)\leq\lambda(x;u)=\lambda(x;u+v-v)\leq\max\{\lambda(x;u+v),\lambda(x;v)\}$$ If $\lambda(x;u+v)\leq\lambda(x;v)$, then $\lambda(x;u)\leq \lambda(x;v)$ which leads to a contradiction. Therefore $\lambda(x;u+v)> \lambda(x;v)$ and so \begin{align} \lambda(x;u+v)=\lambda(x;u)=\max\{\lambda(x;v),\lambda(x;u)\} \tag{2}\label{two} \end{align}

This case is weaker than what the OP states in the posting ( $u+v\neq0$ ). The best I have obtain in that direction is \begin{align} \lambda(x;u+v)&\leq \max\{\lambda(x,u),\lambda(x,v)\}\\ &=\limsup_n\frac1n\log\big(\sqrt{\|A^n(x)u\|^2+\|A^n(x)v\|^2}\big)\\ &=\limsup_n\frac1n\log\Big(\sqrt{\frac{\|A^n(x)(u+v)\|^2+\|A^n(x)(u-v)\|^2}{2}}\Big)\\ &=\limsup_n\frac{1}{2n}\log\big(\|A^n(x)(u+v)\|^2+\|A^n(x)(u-v)\|^2\big)\\ &=\max\big\{\lambda(x;u+v),\lambda(x,u-v)\big\} \end{align} which follows from the identity \begin{align} \limsup_n\frac1n\log(a_n+b_n)&=\max\big\{\limsup_n\frac1n\log(a_n),\limsup_n\frac1n\log(b_n)\big\}\\ &=\limsup_n\frac1n\log\big(\sqrt{a^2_n+b^2_n}\big) \end{align} for sequences $a_n,b_n>0$.

Edit: As user John B has shown, the statement of the problem as it appears in M. Viana's textbook does not hold in general when $\lambda(x,u)=\lambda(x,v)$.