Given are a continuous, strictly increasing and convex function $\varphi:[0,\infty)\to[0,\infty)$ and a sequence $(x_j)$ of positive real numbers such that \begin{align*} \sum_{j=1}^\infty\varphi(x_j) \end{align*} converges.
Question: Can we always find a sequence $(\lambda _j)$ of positive real numbers which decreases to 0 and so that \begin{align*} \sum_{j=1}^\infty\varphi(x_j/\lambda_j) \end{align*} still converges?
This is true for linear $\varphi$ by the Abel-Dini-Theorem, and in this case one can take $\lambda_j=\sqrt{x_{j+1}+x_{j+2}+\ldots}$. But what about the general case? Any help is highly appreciated. Thanks in advance!
No - here is a counterexample. Define
$$\phi(x)=\begin{cases} 0&x=0\\ \exp(-1/x)&0<x\leq 1/2\\ \exp(-2)+4\exp(-2)(x-\tfrac12)&x\geq 1/2. \end{cases}$$
For $0<x<1/2$ we have $\phi'(x)=x^{-2}\exp(-1/x)$ and $\phi''(x)=(x^{-4}-2x^{-3})\exp(-1/x)>0$ which ensures that $\phi$ is continuous, strictly increasing and convex.
For $0<\lambda\leq 1$ and $0<y^\lambda<\exp(-2)$ we have
$$\phi(\tfrac1\lambda \phi^{-1}(x))=\phi(-1/\lambda\log(y))=y^\lambda.$$
For an extreme example take $x_j=\phi^{-1}(1/j(\log j)^2)$ for $j\geq 10$ (and $x_1,\dots,x_9=1$ say). Then
$$\sum_{j=10}^\infty \phi(x_j/\lambda)=\sum_{j=10}^\infty \frac{1}{j^\lambda(\log j)^{2\lambda}}$$
which converges for $\lambda=1$ but not for any $\lambda<1.$