Convergence of the sum of a family of real-valued functions Let $\phi_1,...,\phi_n,...$ be a sequence of real-valued functions so that $\phi_j:[0,1)\to[0,1)$, $\phi_j(0)=0$, and $\phi_j(\delta)$ converges to 0 as $\delta$ approaches 0 from the right for all $j\ge1$. Further suppose that $\sum_{j=1}^\infty \phi_j(\delta)$ converges and moreover, is strictly smaller than $1$ for an arbitrary $\delta$ in $(0,1)$.

MY QUESTION is:
Does it hold that $\lim_{\delta\to0^+} \sum_{j=1}^\infty \phi_j(\delta)=0$  ? If it doesn't, then does it turn out to be true if we in addition require that $\lim_{\delta\to0^+}{\phi_j(\delta) \over \delta}=0$ for all $j\ge1$?

The sequence satisfying $\phi_j(\delta)=\delta^j$ for all $j\ge1$, which gave me some kind of intuition, does support the equality, but I guess the equality does not hold true in general and I've been stuck in disproving it for days... Now I really need someone to show me a complete and detailed proof or disproof of the equality... Any help would be appreciated!
 A: Let $\phi_i'(x)$ be defined as $1/2$ for $1/2^i<x<1/2^{i-1}$ and 0 elsewhere.
Clearly $\phi_j':[0,1)\to[0,1)$, $\phi_j'(0)=0$, and $\phi_j'(\delta)$ converges to 0 as $\delta$ approaches 0 from the right for all $j\ge1$. Also $\sum_{j=1}^\infty \phi_j'(x)$ converges and is equal to 0 if $x=0$ and $1/2$ elsewhere.
This is therefore a counterexample to your assertion since $\lim_{\delta\to0^+} \sum_{j=1}^\infty \phi_j(\delta)=1/2 \neq 0$.
It also shows that adding the further condition $\lim_{\delta\to0^+}{\phi_j(\delta) \over \delta}=0$ for all $j\ge1$ doesn't help since each $\phi_i'$ is identically $0$ in $[0,1/2^i)$ which is a right neighbourhood of $0$.
An obvious condition that does work is to assert that the series $\sum_{j=1}^\infty \phi_j'(x)$
converges uniformly at $0$. This is because $\phi_i$ is continuous at $0$ and if $\sum_{j=1}^\infty \phi_j'(x)$ also converges uniformly at 0 then it must also be continuous at $0$ and hence $\lim_{\delta\to0^+} \sum_{j=1}^\infty \phi_j(\delta)=\sum_{j=1}^\infty \phi_j(0)=0$.
