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I am reading Arnold's ODE (3rd edition) and I am confused with the remark.

Page 62 :

Remark. The condition of smooth dependence on the time t is needed in order to eliminate pathological examples such as the following:

let { $\alpha$ } be a basis of the group R i.e., a set of real numbers such that each real number has a unique representation in the form of a finite linear combination of numbers of the set with integer coefficients.

Q1.) I think he is not talking about the vector basis since the integer is not even a field.

How can I show there indeed exists such "basis" ?

To each number $\alpha$ of the basis we assign the translation of the line by some distance, paying no attention to other elements of the basis. Setting $g^{\alpha_1+...+\alpha_k}=g^{\alpha_1}...g^{\alpha_k}$, we obtain a one-parameter transformation group each of whose element is a translation of the line consequently a differmorphism; but in general $g^t$ is not a smooth function of t and is even discontinuous.

Q2.) I don't know how this reasoning shows $g^t$ must be smooth with respect to t, as he remarks at the beginning.

Instead of smoothness with respect to t one may require only continuous(from which smoothness is a consequence) but we have no need to do this.

Q3.) I don't understand this either. Please give me a hint.

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Math The Novice
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Regarding Q1), I think that you cannot show anything, because such a basis does not exist. Indeed, suppose it exists, and pick some (necessarily nonzero) $\alpha$ belonging to it. $\alpha/2$ is a nonzero real number, so it must be uniquely expressed in the form $$ \frac{\alpha}{2} = k_1 \alpha_1 + \ldots + k_n \alpha_n, $$ where $k_1, \dots, k_n \in \mathbb{Z} \setminus \{0\}$, $\alpha_1, \dots, \alpha_n$ belong to the basis. Then we have $$ \alpha = (2k_1) \alpha_1 + \ldots + (2k_n) \alpha_n, $$ which means that $\alpha$ has two representations, a contradiction.

I would look for a such a pathological one-parameter group in the following way. Take any non-measurable solution of Cauchy's functional equation $h(t + s) = h(t) + h(s)$, and define $$ g^t(x) = e^{h(t)}x, \quad x \in \mathbb{R}. $$

EDIT: We have obtained a homomorphism of the additive group of reals into the groups of scalings of $\mathbb{R}$. I chose it since it looks to me more natural in the context of linear differential equations. However, to remain in the setup of the textbook one should rather choose a homeomorphism of the additive group of reals into the groups of translations of $\mathbb{R}$. Then $$ g^t(x) = x + h(t), \quad x \in \mathbb{R}. $$

Now, you can find answers to Q2) and Q3) in Overview of basic facts about Cauchy functional equation.

user539887
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    I still don't get Q3). I am not sure how to interpret the group action $g^{t+s}=g^t \circ g^s$ as a Cauchy functional equation. In the $f:\mathbb{R} \to \mathbb{R}$ case, the link you give says $f(t) = e^{ct}$ if $f(t)$ is continuous, but how do I interpret this $e^{ct}$ in group action? – Math The Novice Aug 10 '18 at 15:13
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    @MathTheNovice I edited the answer: gave a different example of a pathological one-parameter group. – user539887 Aug 10 '18 at 18:52
  • Thanks. However, I want to ask $g^{t+s}=g^t \circ g^s$ whether continuity $g^t$ w,r,t t automatically implies smoothness w.r.t t in general. (For the example mentioned, Yes. But this seems to be false in general, I am not sure.) – Math The Novice Aug 11 '18 at 05:39
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    @MathTheNovice On the real line, yes, continuity implies that $g^t=e^{ct}$ for some $c\in\mathbb{R}$ (I need to find all functions f:R→R which are continuous and satisfy f(x+y)=f(x)+f(y)), and the latter is differentiable. Perhaps you mean a one-parameter (semi)group of linear operators on a Banach space? – user539887 Aug 11 '18 at 07:55
  • Yes. Is "the continuity with respect to t implies smoothness" true? – Math The Novice Aug 12 '18 at 07:41
  • @MathTheNovice No, in general this is not true (semigroups satisfying that for $t>0$ are called differentiable, see C0-semigroup). There is one essential difference, however: in the case of reals, finding "pathological" examples requires applying the Axiom of Choice, whereas in infinite-dimensional Banach spaces one explicitly gives a construction. – user539887 Aug 12 '18 at 08:56