I am reading Cartan's original paper
Sur les mesures de Haar, comptes rendus de l'Académie des Sciences de Paris, 1940
on the existence of Haar measure on a locally compact Hausdorff topological group. There is a lemma I'm stuck on :
Translation :
Given finitely many $f_i$ continuous nonnegative with compact support, $\rho > 0$ and $\Lambda > 0$, there is a neighborhood $U$ of the identity in $G$ such that $$ I_{\phi}\bigg(\sum_i \lambda_i f_i\bigg) \le \sum_i \lambda_iI_{\phi}(f_i) \le I_{\phi}\bigg(\sum_i \lambda_i f_i\bigg) + \rho $$ whenever $\phi \neq 0$ is continuous, nonnegative with compact support in $U$, and $0 \le \lambda_i \le \Lambda$.
The functional $I_\phi$ is defined by $$ I_{\phi}(f) = \frac{(f:\phi)}{(f_0 : \phi)} $$ with usual notation for Haar covering numbers and $f_0$ is a fixed nontrivial nonnegative continuous function with compact support in $G$.
This result is well-known in the particular case $\lambda_i=1$ for all $i$. For example, we can find a proof in Folland's Real analysis, or in André Weil's book L'intégration sur les groupes topologiques et ses applications (page 36) as mentionned by Cartan in the footnote.
Later on in the proof of the Théorème d'approximation, Cartan defines some $\lambda_i$ depending on $\phi$, so the $U$ given by the lemma must be the same for all possible choices of $\lambda_i \le \Lambda$ (I thought at first that it was a quantifier misordering, but it would be a surprising mistake, coming from Cartan). However, I don't know, after many many attempts, how to prove this lemma in its « $\lambda_i-$uniform formulation ».
Is it even possible to deduce it from the $\lambda_i = 1$ case ?
Any hint will be appreciated. :)
I think that I have a solution.
Pick a compact neighborhood $V$ of the identity such that $\operatorname{supp} f_i \subset V$ for all $i=1,\ldots,n$.
Let $K = V V$, which is compact, and choose $g$ continuous with compact support such that $$ 1_K \le g \le 1$$
Let $\delta > 0$ be given. Define for any $i=1,\ldots,n$ and $\lambda = (\lambda_1,\ldots,\lambda_n) \in [0,\Lambda]^n$ a function $$ h_{i,\lambda} = \frac{\lambda_i f_i}{ \delta g + \sum_{j=1}^n \lambda_j f_j} 1_V \in C_c(G) $$ These are continuous, nonnegative, with compact support and $\lambda_i f_i = \big(\delta g +\sum_j \lambda_j f_j\big) h_{i,\lambda} $.
The mappings $\lambda \mapsto h_{i,\lambda}|_K$ are continous from $[0,\Lambda]^n$ to $C(K)$, the space of continuous functions on $K$. Therefore the family $\mathcal H = \{h_{i,\lambda} : \lambda \in [0,\Lambda]^n \text{ and } i=1,\ldots,n\}$ is compact in $C(K)$. According to the Arzelà-Ascoli theorem, $\mathcal H$ is equicontinuous. Moreover, as $K$ is compact and inherits the (left and right) uniform structures on $G$, $\mathcal H$ is actually (left and right) uniformly equicontinuous.
So there is some symmetric neighborhood $U$ of the identity such that $U\subset V$ and $$ \tag{$*$} \forall h \in \mathcal H,\forall x,y\in K~,~ y^{-1}x \in U \Longrightarrow |h(x)-h(y)| \le \delta $$ Let $\phi \neq 0$ be nonnegative, continuous with compact support in $U$, and let $0\le \lambda_1,\ldots,\lambda_n \le \Lambda$.
Suppose that finitely many $s_k \in G$ and $c_k \ge 0$ are given so that $$ \forall x\in G ~,~ \delta g(x) + \sum_j \lambda_j f_j(x) \le \sum_k c_k \phi(s_k^{-1}x) $$ For any $i=1,\ldots,n$ we have $$\forall x\in G ~,~ \lambda_i f_i(x) \le \sum_k c_k h_{i,\lambda}(x) \phi(s_k^{-1}x) $$ Take $x$ in the support of $f_i$. Either $s_k^{-1} x \in U$ : by symmetry of $U$, $s_k \in xU \subset K$ thus by $(*)$ $$ h_{i,\lambda}(x)\phi(s_k^{-1}x) \le \big(h_{i,\lambda}(s_k) + \delta\big)\phi(s_k^{-1}x) $$ Or $s_k^{-1}x\in U$ and the above inequality still holds as $\operatorname{supp} \phi \subset U$. Hence $$\lambda_i f_i(x) \le \sum_k c_k \big(h_{i,\lambda}(s_k) + \delta\big) \phi(s_k^{-1}x)$$ This inequality remains valid for any $x$ in $G$. By definition of the Haar covering numbers : $$ \lambda_i (f_i: \phi) \le \sum_k c_k \big(h_{i,\lambda}(s_k) + \delta\big) $$ Summing up over $i$, using $\sum_i h_{i,\lambda} \le 1$, we get $$ \sum_i \lambda_i (f_i : \phi) \le (1+n\delta) \sum_k c_k $$ Since the $s_k$ and $c_k$ were arbitrary, we have $$ \begin{split} \sum_i \lambda_i (f_i : \phi) &\le (1+n\delta) \delta (g:\phi) + (1+n\delta)\bigg(\sum_i \lambda_i f_i : \phi\bigg)\\ &\le \bigg(\sum_i \lambda_i f_i : \phi\bigg) + \delta(1+n\delta)(g:\phi) + n\delta \Lambda\bigg(\sum_i f_i : \phi\bigg) \end{split}$$Divide both sides by $(f_0 : \phi)$ $$ \begin{split} \sum_i \lambda_i I_\phi(f_i) &\le I_\phi\bigg(\sum_i \lambda_i f_i\bigg) + \delta(1+n\delta)(g:f_0) + n\delta \Lambda\bigg(\sum_i f_i : f_0\bigg) \end{split}$$ Once we have chosen $\delta$ small enough to satisfy $$ \delta(1+n\delta)(g:f_0) + n\delta \Lambda\bigg(\sum_i f_i : f_0\bigg) \le \rho $$ the proof of Cartan's lemma is complete.