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Is it true for a (upper/lower) semicontinuous function $f$ and a monotonically increasing function $g$ which is also (upper/lower) semicontinuous) that $f\circ g$ is also (upper/lower)semicontinuous?

Somehow I think that this is true( just a Feeling), but I don't know which equivalent definition of semicontinuity I can use to prove this?

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    $\begingroup$ Didn't you mean $g \circ f$? $\endgroup$ Jan 20, 2014 at 11:20
  • $\begingroup$ Actually no, but if it is true for $g \circ f$, then this would be useful too. Can you give me a hint, how to prove it for $g \circ f$? $\endgroup$
    – user66906
    Jan 20, 2014 at 11:31
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    $\begingroup$ Hints: 1) for $f\circ g$ find counterexample when $f$ is continuous; 2) for $g \circ f$ apply $g$ to both sides of the inequality in the definition of semicontinuous function $\endgroup$
    – Skeeve
    Jan 20, 2014 at 11:47
  • $\begingroup$ regardubg 2.) let's assume lsc then f is lsc @$x_0$ if we have that $f(x)+\epsilon \ge f(x_0)$. applying $g$ we get $g(f(x)+\epsilon) \ge g(f(x_0))$. What we want to have is $g(f(x)) + \epsilon \ge g(f(x_0))$, but this is true, since $g$ itself is lsc and this means $g(y) + \varepsilon \ge g(y + \epsilon)$, right? $\endgroup$
    – user66906
    Jan 20, 2014 at 13:15
  • $\begingroup$ @Lipschitz well, $g$ in general is not lsc $\endgroup$
    – Skeeve
    Jan 20, 2014 at 22:00

1 Answer 1

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Claim 1. If $f$ and $g$ are lower semicontinuous and $g$ is monotone then $f\circ g$ can fail to be lower semicontinuous.

Indeed, take $f(x)=-x$ and $g(x)=\begin{cases}x & x\le 0 \\ x+1 & x>0\end{cases}$. Then $f\circ g$ is not lower semicontinuous.

Note. In this example $f\circ g$ is however upper semicontinuous. But one can make $f\circ g$ worse by taking more complicated $f$, for instance $f(x)=\begin{cases}-x & x \le 1 \\ \sin\frac{1}{x-1} & x>1\end{cases}$. In this case $f\circ g$ is neither lower nor upper semicontinuous. (Observe that the graph of $f\circ g$ is obtained from the graph of $f$ essentially by "cutting" away $(0,1]$ from $x$-axis...)

Claim 2. If $f$ and $g$ are lower semicontinuous and $g$ is monotone then $g\circ f$ is lower semicontinuous.

Let us fix $x_0$.

By lower semicontinuity of $g$ for any $\varepsilon>0$ there exists $\delta>0$ such that $g(f(x_0)-\delta) \ge g(f(x_0))$

Now, by lower semicontinuity of $f$ there exists $\gamma >0$ such that $f(x)\ge f(x_0) - \delta$ if $|x-x_0|<\gamma$. Using monotonicity of $g$ we get $g(f(x)) \ge g(f(x_0)- \delta)$. And using the inequality we obtained earlier we get $g(f(x)) \ge g(f(x_0))$, hence $g\circ f$ is lower semicontinuous.

Note. If $g$ is just monotone, Claim 2 does not hold. Just take any monotone not semicontinuous function $g$ and $f(x)=x$. (Sorry if my comments were confusing)

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  • $\begingroup$ okay, thank you. then I agree with you. $\endgroup$
    – user66906
    Jan 21, 2014 at 16:43

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