# Convergence of a series given properties on uniformly bounded functions

I have the following problem from "Bollobas, Linear analysis. An introductory course"

Let $$\phi_n:[0,1]\to\mathbb{R}^+$$ $$(n=1,2,\dots)$$ be uniformly bounded continuous function such that $$\int_{0}^1\phi_n(x)dx\geq c$$ for some $$c>0$$. Suppose $$c_n\geq 0$$, $$(n=1,2,\dots)$$ and $$\sum_{n\geq 1}c_n\phi_n(x)<\infty$$ for every $$x\in[0,1]$$. Prove that $$\sum_{n\geq 1}c_n<\infty$$

I don't know how to proceed. Any hints will be useful !

• Really the only traction you have on $\phi_n$ is its integral on $[0,1]$. Have you tried integrating your $c_n \phi_n$ sum over $[0,1]$ to see what happens? – Eric Towers Dec 6 '18 at 2:42
• @EricTowers It is not given that $\sum c_n \phi_n$ is integrable. – Kavi Rama Murthy Dec 6 '18 at 5:38
• @KaviRamaMurthy : Lebesgue tells me jnaf can (Riemann) integrate each $c_n\phi_n$. I would wonder if jnaf can justify swapping the order of integration and summation, but thinking about this aspect reminds me that $\phi_n(x) \in \ell^\infty$ for each $x$ and the $\phi_n$ are uniformly bounded away from zero. – Eric Towers Dec 6 '18 at 13:53
We know that the sequence $$(\varphi_n)_{n \in \mathbb{N}}$$ is uniformly bounded, i.e. we have $$|\varphi_n|_\infty \le M$$ for all $$n \in \mathbb{N}$$ for some $$M>0$$ independent of the index $$n$$. Moreover, we write $$B_n:= \{\varphi_n \ne 0 \}$$ and let $$X := \bigcup_{n=1}^\infty B_n.$$ Note that $$X$$ is open and not empty. In fact, we have $$B_n \neq \emptyset$$ for all $$n \in \mathbb{N}$$, because of $$\int_0^1 \varphi_n(t) \, dt \ge c$$. In particular, $$0< \lambda(X) \le 1$$. (Moreover, $$X$$ is a Baire-space as an open subset of a complete metric space. But we will not need this.)
Define $$h(x) := \sum_{k=1}^\infty c_k \varphi_k(x).$$ This series is pointwise convergent, as assumed, and therefore also measurable. Let $$A_n := \{x \in X : n < h(x) \le n+1\}$$ and note that $$X = \bigcup_{n=1}^\infty A_n$$. By $$\sigma$$-additivity of the measure, we have $$\lambda(X) = \sum_{k=1}^\infty \lambda(A_k).$$ Hence the series is convegent and we find for some $$N \in \mathbb{N}$$ that $$\sum_{k=N}^\infty \lambda(A_k) < \frac{c}{2M}.$$ Set $$Y:= \bigcup_{k=1}^N A_k$$. Note that for any $$x \in Y$$ we have $$|h(x)| \le N+1$$ and $$\lambda(X \setminus Y) < c/(2M)$$. Therefore we get $$\int_Y \varphi_n(x) \,dx \ge c- \int_{X \setminus Y} \varphi_n(x) \,dx \ge \frac{c}{2}.$$ The monotone convergence theorem implies now $$\frac{c}{2} \sum_{n=1}^\infty c_n \le \sum_{k=1}^\infty c_k \int_Y \varphi_k(x) \, dx = \int_Y \sum_{k=1}^\infty c_k \varphi_k(x) \, dx = \int_Y h(x) \, dx \le N+1.$$ Thus the series in the last equation is convergent.