# Nonlinear operator sends bounded set to relatively compact set

Consider $g$ a continuous function on $[a,b]\times\mathbb{R}$, and let $z_0\in\mathbb{R}$. Define the (nonlinear) operator on $C[a,b]$: $$Mv(x)=z_0+\int_a^x g(t,v(t))\,dt$$ for $x\in[a,b]$. Prove that:

(i) $M$ is continuous and

(ii) maps bounded set in $C[a,b]$ to relatively compact set (closure is compact) in $C[a,b]$.

My attempt: I found difficulty in this question since the operator is nonlinear, so most of the theorems on functional analysis are not applicable since they are only for linear operators.

For continuity of $M$, I let $(v_n)$ be a sequence in $C[a,b]$ convering to $v\in C[a,b]$. Then show that $g(t,v_n(t))$ converges to $g(t,v(t))$ uniformly. Then, I show that $\lim_{n\to\infty}Mv_n(x)=Mv(x)$, so $M$ is continuous (sequential criteria for continuity).

For the second part, I am facing difficulties. I have an idea to use Arzela-Ascoli Theorem (Generalised form): "Let $X$ be a compact Hausdorff space. Then a subset $F$ of $C(X)$ is relatively compact in the topology induced by the uniform norm if and only if it is equicontinuous and pointwise bounded." However I have problems showing equicontinuous and pointwise bounded.

Thanks for any hints!

For the second part, take a bounded subset $B$ of $C[a,b]$. We can find a constant $R$ such that $|v(x)|\leqslant R$ for each $x\in [a,b]$ and each $v\in B$. To prove that $M(B)$ is equi-continuous and bounded, use the fact that $g$ is uniformly continuous (and bounded) on $[a,b]\times [-R,R]$.
• How do we ensure the same $M$ works for all the $v\in B$? – yoyostein Feb 7 '16 at 15:56