Show that the function $h$ defined by $h(x)=g(f(x))$ is lower semicontinuous.

Let $$f:\mathbb{R}^n\longrightarrow \mathbb{R}$$ be a lower semicontinuous function, and $$g:\mathbb{R}\longrightarrow\mathbb{R}$$ be a lower semicontinuous and nondecreasing function.

(1) Show that the function $$h$$ defined by $$h(x)=g(f(x))$$ is lower semicontinuous.

(2) Give an example showing that the nondecrease assumption is essential.

Part (1)
$$f$$ is lower semicontinuous, therefore: $$\forall x_0\in\mathbb{R}^n$$

$$\forall\epsilon>0: \exists U\in N(x_0): f(x)\geq f(x_0)-\epsilon,\; \forall x\in U \tag{1}$$ Where $$U$$ is a neighborhood of $$x_0$$.

$$g$$ is lower semicontinuous, therefore: $$\forall y_0\in\mathbb{R}$$

$$\forall\epsilon>0: \exists V\in N(y_0): g(y)\geq g(y_0)-\epsilon,\; \forall y\in V \tag{2}$$

It is sufficient to prove for $$h(x)=g(f(x))$$ that: $$\forall x_0\in\mathbb{R}^n$$

$$\forall\epsilon>0: \exists U\in N(x_0): g(f(x))\geq g(f(x_0))-\epsilon,\; \forall x\in U \tag{3}$$

Attempt
$$g$$ is nondecreasing, therefore $$(1)$$ becomes: $$\forall x_0\in\mathbb{R}^n$$

$$\forall\epsilon>0: \exists U\in N(x_0): g(f(x))\geq g(f(x_0)-\epsilon),\; \forall x\in U \tag{4}$$

I am stuck here. I am also looking at the possibility of taking $$y_0=f(x_0)$$ and $$y=f(x)$$.

I am frankly lost in the abstraction, and I do not fully understand why the lower semicontinuity is important here, or why $$g$$ must be nondecreasing.

Part (2)

I feel if I understand Part (1) then it would be easier for me to come up with an example.

Thank you.

It might help to work with the sequential characterisation of lsc. A function $$h$$ is lsc. at $$x$$ iff $$\liminf_n h(x_n) \ge h(x^*)$$ for any sequence $$x_n \to x^*$$.
Pick some sequence $$x_n \to x^*$$. Since $$f$$ is lsc. we have $$\liminf_n f(x_n) \ge f(x^*)$$, and since $$g$$ is non decreasing we have $$g(\liminf_n f(x_n) ) \ge g(f(x^*))$$.
We have $$\lim_n \inf_{k \ge n} f(x_k) = \liminf_n f(x_n)$$ (the definition) and so $$\liminf_n g(\inf_{k \ge n} f(x_k)) \ge g(\liminf_n f(x_n)) \ge g(f(x^*))$$.
Since $$g$$ is non decreasing and $$f(x_n) \ge \inf_{k \ge n} f(x_k)$$ we have $$g(f(x_n)) \ge g(\inf_{k \ge n} f(x_k))$$, and so $$\liminf_n g(f(x_n)) \ge \liminf_n g(\inf_{k \ge n} f(x_k)) \ge g(f(x^*))$$ and so $$g \circ f$$ is lsc.
As an example of why the non decreasing is necessary, we can take a continuous $$g(x) = -x$$ and let $$f = 1_{(0,\infty)}$$ (indicator function of the strictly positive reals). Then with $$x_n = {1 \over n}$$ we have $$f(x_n) = 1$$ and $$\liminf_n f(x_n) = 1 \ge f(0) = 0$$, but $$g(f(x_n)) = -1$$ and $$\liminf_n g(f(x_n)) = -1$$, but $$g(f(0)) = 0 > -1$$, so $$g \circ f$$ is not lsc.
• In the last sentence, it should say "but $g(f(x_n))=-1$". May 31 '19 at 13:04