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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.

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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.

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  • $\begingroup$ In the last sentence, it should say "but $g(f(x_n))=-1$". $\endgroup$ May 31 '19 at 13:04
  • $\begingroup$ @EverydayAstronaut: Thanks! $\endgroup$
    – copper.hat
    May 31 '19 at 13:11

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