Let $ f:\mathbb{R}^m \rightarrow (-\infty,\infty] $ be lower semicontinuous and bounded from below. Set $f_k(x) = \inf\{f(y)+k d( x,y ): y\in \mathbb{R}^m\} $ , where $d(x,y)$ is a metric. It is easy to see that each $f_k$ is continuous and $f_1 \leq f_2\leq …\leq f \\$. However, I don't know how to prove that $ \lim_{k \rightarrow \infty}f_k(x) = f(x) $ for every $x\in\mathbb{R}^m $.
[Math] Lower semicontinuous function as the limit of an increasing sequence of continuous functions
real-analysissemicontinuous-functions
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This is a quotation from "General Topology" by Ryszard Engelking:
[Math] Composition of lower semicontinuous function with continuous function is lower semicontinuous
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.
Best Answer
Sea $-M$ una cota inferior de $f$ (con $M > 0$).\ Tomemos $x\in X$, para todo $\epsilon > 0$ existe $\delta > 0$ tal que si $y\in B(x,\delta)$ entonces $f(y) > f(x) - \epsilon$.\ Por otro lado, para todo $k\geq 1$ existe $y_k \in X$ tal que
$$f(y_k) + kd(x,y_k) < f_k(x) + \epsilon \leq f(x) + \epsilon$$
Como $f(y_k) \geq f_k(x) \geq -M$ la desigualdad $kd(x,y_k) < f(x) + \epsilon + M$ es satisfecha. Luego, dividiendo por $k$ se tiene que $0\leq d(x,y_k) < \dfrac{(f(x) + \epsilon + M)}{k}$, como esto sucede para todo $k\geq 1$ deducimos que $d(x,y_k)$ converge a $0$. Luego existe $k_0 \geq 1$ para el cual $d(x,y_k) < \delta$ cuando $k\geq k_0$. De esto, se cumple que $f(y_k) > f(x) - \epsilon$ para todo $k\geq k_0$.\ Luego $0\leq f(x) - f_k(x) < f(x) - (f(y_k) + kd(x,y_k) - \epsilon) = (f(x) - f(y_k)) - kd(x,y_k) + \epsilon < 2\epsilon$ para todo $k\geq k_0$. Por lo tanto $f_k(x)$ converge a $f(x)$.