Assume that $f\colon \mathbb{R}^n \to\mathbb{R}$ is lower semicontinuous at $g(a)$ and $g\colon \mathbb{R}^m \to\mathbb{R}^n$ is continuous at $a \in\mathbb{R}$. Define $h = f \circ g \colon \mathbb{R}^m \to \mathbb{R}$ by $h(x) = f(g(x))$ for every $x \in \mathbb{R}^n$. Then prove that $h$ is lower semicontinuous at $a$.
I am unsure how to start this proof, any suggestions$?$
Best Answer
According to the definition you provide in the comments, to prove that $h$ is lower semicontinuous at $a$ we need to prove that, for any $\varepsilon >0$ there exists $\delta >0$ such that $h(x)>h(a)-\varepsilon$, when $\|x-a\|<\delta$.
We know, since $g$ is continuous at $a$, that for any given $\varepsilon>0$ there exists $\delta_1>0$ such that $\|g(x)-g(a)\|<\varepsilon$, if $\|x-a\|<\delta_1$.
Because $f$ is lower semicontinuous at $g(a)$, we know that, for any $\varepsilon >0$, there exists $\delta_2>0$ such that $f(y)>f(g(a))-\varepsilon$, when $\|y-g(a)\|<\delta_2$.
Let $\varepsilon>0$. By the fact that $f $ is lower semicontinuous at $g(a)$, by only considering $y=g(x)$ (that is, taking $y$ in the range of $g$) we know that there exists a $\delta_2>0$ such that $\|g(x)-g(a)\|<\delta_2$ implies $f(g(x))>f(g(a))-\varepsilon$. Now, by choosing $\varepsilon=\delta_2$ in the definition of continuity for $g$, we get that there exists $\delta_1>0$ such that if $\|x-a\|<\delta_1$ then $\|g(x)-g(a)\|<\delta_2$.
By putting the two together and taking $\delta=\delta_1$, we get that for any $\varepsilon>0$ there exists $\delta>0$ such that $f(g(x))>f(g(a))-\varepsilon$ when $\|x-a\|<\delta$, which is what we wanted to prove.