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Hi does anyone have any idea or a possible hint for a proof of the following result:

Consider asymmetric norm $p$ on $\mathbb{R}$ given by $p(t) = t^{+}$, for $t \in \mathbb{R}$. Show that if $(X, \tau)$ is a topological space, then a real-valued function $f$ on $(X, \tau)$ is upper-semicontinuous as a function from $(X, \tau)$ to $(\mathbb{R}, |\cdot|)$ iff it is continuous from $(X, \tau)$ to $(\mathbb{R},p)$.

Thanks for any assistance.

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  • $\begingroup$ $f$ is upper semicontinuous if and only if $f^{-1}([c,+\infty[)$ is closed for all $c\in \mathbb{R}$. How is the topology induced by $p$ defined? Is the family $\{ y : p(y-x) < \varepsilon\}$ for $\varepsilon > 0$ a neighbourhood basis of $x$? $\endgroup$ – Daniel Fischer Jun 23 '15 at 12:25
  • $\begingroup$ @DanielFischer Yes the topology is defined from the induced metric as you stated. $\endgroup$ – Lucio D Jun 23 '15 at 12:44
  • $\begingroup$ In that case, just look what the open resp. closed sets in that topology are. Given $x$ and $\varepsilon$, what is $\{ y : p(y-x) < \varepsilon\}$? $\endgroup$ – Daniel Fischer Jun 23 '15 at 12:48
  • $\begingroup$ @DanielFischer Okay, is my proof of the first implication fine? $\endgroup$ – Lucio D Jun 23 '15 at 13:41
  • $\begingroup$ Hmm. At this stage, I would demand a little more explicit argument why "$f^{-1}((-\infty,x+\varepsilon))$ is open in $(X,\tau)$" implies the continuity of $f$ with respect to the topology induced by $p$. Since you are new to this stuff, I wouldn't like to assume that you have understood it based on an abridged sketch of the argument. $\endgroup$ – Daniel Fischer Jun 23 '15 at 14:00
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For the first implication (=>)

Assume that $f: (X,\tau) \rightarrow (\mathbb{R}, |\cdot|)$ is upper semi-continuous. Let $\{y: p(y-x) < \epsilon \}$ be a basic open neighbourhood of $x$ in $(\mathbb{R}, p)$, then since $$ p(y-x) = \begin{cases} y-x ~~~~\text{if}~~~~~y \geq x, \\ 0~~~~~~~~~~~\text{if}~~~~~y<x \end{cases} $$ it follows that $\{y: p(y-x) < \epsilon \} = \{y:y-x < \epsilon\} = (-\infty, x+ \epsilon )$.

As you stated "$f$ is upper semi-continuous if and only if $f^{-1}([c,\infty))$ is closed for all $c \in \mathbb{R}$."

We have that $[x + \epsilon, \infty) = \mathbb{R} \setminus (-\infty, x + \epsilon)$. Since $f^{-1}(\mathbb{R} \setminus (-\infty,x+\epsilon)) = X \setminus f^{-1}((-\infty, x + \epsilon))$ is closed in $X$, it follows that $f^{-1}((-\infty,x+\epsilon))$ is open in $X$.

Therefore $f: (T,\tau) \rightarrow (\mathbb{R},u)$ is continuous.

The second implication (<=)

The basic open neighbourhoods are of the form $B(x,\epsilon) = (-\infty, x + \epsilon)$, for some $x \in U$ and $\epsilon > 0$. Since an open set is the union of basic open neighborhoods contained in $U$. The open sets in $(\mathbb{R},u)$ are of the form $(-\infty,a)$. Since we assume that $f: (X, \tau) \rightarrow (\mathbb{R}, u)$ is continuous, it follows that $f^{-1}\big( (-\infty, a) \big)$ is open in $(\mathbb{R}, u)$. But we know the characteristic that $f: (X,\tau) \rightarrow (\mathbb{R},u)$ is upper semi-continouous iff $f^{-1}\big( (-\infty,a) \big)$, where $a \in \mathbb{R}$, is open in $X$. Thus the result is shown.

$\square$

The definition (as stated in wiki entry) of $\limsup\limits_{x \rightarrow x_{0}}f(x)$ that I am using is $$\limsup\limits_{x \rightarrow x_{0}}f(x)=\lim\limits_{\epsilon \rightarrow 0}\big(\sup\{f(x):x \in B(x_{0},\epsilon)\setminus\{x_{0}\}\big).$$ To prove the first implication of the characterization of semi-continuity with your definition, my proof is as follows:

Assume $f(x_{0}) < c$ for some $c \in \mathbb{R}$. Then $x_{0} \in f^{-1}\big((-\infty,c)\big)$, this implies that there is an $\epsilon > 0$ and an open ball $B(x_{0},\epsilon)$ such that $B(x_{0},\epsilon) \subset f^{-1}\big((-\infty,c)\big)$. Therefore for any $y \in B(x_{0},\epsilon) \setminus \{x_{0} \}$, it follows that $f(y) < c$. Hence, $\sup \big\{f(y): y \in B(x_{0},\epsilon) \setminus \{ x_{0} \} \big\} < c$.

I also observe that for any $0 < \epsilon_{1} < \epsilon$ we have $$\sup\big\{f(y) : y \in B(x_{0}, \epsilon_{1}) \setminus \{x_{0}\}\big\} < \sup\big\{f(y) : y \in B(x_{0}, \epsilon) \setminus \{x_{0}\}\big\} < c.$$

I am not sure why it is guaranteed that the limit $$\lim\limits_{\epsilon \rightarrow 0}\big( \sup \big\{f(x): x \in B(x_{0}, \epsilon) \setminus \{ x_{0} \}\big\} \big)$$ exists? This is needed to complete the proof.

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    $\begingroup$ From "$f(y) < c$ for all $y \in B(x_0,\epsilon)\setminus \{x_0\}$, you only get $\sup \bigl\{ f(y) : y \in B(x_0,\epsilon)\setminus \{x_0\}\bigr\} \leqslant c$. I should have stated the criterion with a weak inequality in the consequent, $f(x_0) < c \implies \limsup\limits_{x\to x_0} f(x) \leqslant c$, that would have been a little more convenient for the proof. With the strict inequality, you need to pick a $c_1$, $f(x_0) < c_1 < c$. Then $\limsup f(x) \leqslant c_1 < c$. $\endgroup$ – Daniel Fischer Jun 25 '15 at 12:21
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    $\begingroup$ To answer your question at the bottom: the function $$s(\epsilon) := \sup \bigl\{ f(x) : x \in B(x_0,\epsilon)\setminus \{x_0\}\bigr\}$$ is monotonically increasing (non-strictly); $0 < \epsilon_1 < \epsilon \implies s(\epsilon_1) \leqslant s(\epsilon)$. For monotonic functions, the one-sided limits exist at all points (possibly being infinite), and here we have $$\lim_{\epsilon \to 0} s(\epsilon) = \inf \{ s(\epsilon) : \epsilon > 0\}.$$ $\endgroup$ – Daniel Fischer Jun 25 '15 at 12:21
  • $\begingroup$ @DanielFischer Okay I see, thanks. So would your global characterization of upper semi-continuity then read: $f$ is upper semi-continuous iff for all $x_{0} \in X$ and $c \in \mathbb{R}$ such that $f(x_{0}) < c \implies \limsup\limits_{x \rightarrow x_{0}}f(x) \leq c$? Also, do you know of a standard example of a monotonic function with infinite one-sided limit (I'm trying to visualize this)? $\endgroup$ – Lucio D Jun 25 '15 at 13:06
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    $\begingroup$ Yes, it would read that [the $\limsup$ thing is of course upper semicontinuity at $x_0$, so we have "upper semicontinuous $\iff$ upper semicontinuous at each point"]. $\tan \colon \bigl(-\frac{\pi}{2},\frac{\pi}{2}\bigr) \to \mathbb{R}$ is monotonic and has infinite one-sided limits at both end points. Of course, unless we allow $\pm\infty$ as values, a one-sided limit of a monotonic function can only be infinite at the endpoints of the interval of definition. $\endgroup$ – Daniel Fischer Jun 25 '15 at 13:20

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