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Let $f,g:\mathbb{R}\longrightarrow\mathbb{R}$, and assume that $f$ is continuous from the right at $x_0$, and $g$ is continuous from the right at $f(x_0)$.

Is $g\circ f$ continuous from the right at $x_0$?

Intuitively I'm pretty sure this isn't neccessarly right, but I can't think about a counter example

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    Suggestion: take a function $g$ which is *not* continuous from the left at $f(x_0)$, and make $f$ decreasing.2012-01-23

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If $f:x\mapsto-x$, the function $g\circ f$ is continuous from the right, exactly at the points $x$ such that the function $g$ is continuous at $-x$ from the left.

Consider for example $g:x\mapsto[x\geqslant0]$. Then $g$ is continuous at $0$ from the right but not from the left. One can check that $g\circ f:x\mapsto[x\leqslant0]$, hence $g\circ f$ is continuous at $0$ from the left but not from the right.

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    can you explain a bit more please, I'm not sure I understand your answer2012-01-23
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    The function $h=g\circ f$ is defined by $h(x)=g(-x)$. Hence $h(x+\varepsilon)=g(-x-\varepsilon)$. Thus $h(x+\varepsilon)\to h(x)$ when $\varepsilon\to0^+$ iff $g(-x-\varepsilon)\to g(-x)$ when $\varepsilon\to0^+$ if and only if $g$ is continuous at $-x$ from the left.2012-01-23
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    @sony jimbo It might be worthwhile to note that the graph of $g\circ f$ is the graph of $g$ reflected about the $y$-axis (since $\bigl(g\circ f\bigr)(x)=g(-x)\,$).2012-01-23
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    I don't get why $g\circ f:x\mapsto[x\leqslant0]$ is right? won't it map the same way as g?2012-01-23
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    Did you try to compute $g\circ f(x)$ for every $x$ and to draw the graph of $g\circ f$? As said before and recalled by @David, $g\circ f(x)=g(-x)$ hence $g\circ f(x)=[-x\geqslant0]$ hence...2012-01-23
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Well one example that I can think of which is a weird example but it was accepted back in the day when we had the exact same question:

Consider:

$g(x) = \sqrt{x}$ and $f(x) = -|x| $ if $x_0 = 0$ $g(x)$ is continuous to the right and $f(x)$ is continuous in all $\mathbb R$ but $ g\circ f $ is defined only at $x=0$ so at least by our definition of continuity it was not considered continuous at $x_0 = 0$.