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a) If $f$ is $C^{\infty}$ function with compact support, and $g$ is any continous function, $fg$ I guess is not a $C^{\infty}$ function with compact support, but I am not able to produce a counter example.

b) for $g(x)=0$ we get $fg=0$, so this is false

c) Here we need to find a function which blows up at infinity?

please help.

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    ah! thank you for your comment2012-11-01

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OK, a more thorough response: for $(a),$ consider the bump function that's $e^{\frac{-1}{1-x^2}}$ on $(-1,1)$ and $0$ elsewhere, which obviously has compact support and can be shown to be $C^\infty$. Multiply it be $1-|x|$ to get a non-differentiability at $x=0$.

For $(b),$ if $f,g$ has compact support then for any $h$, $fh$ has support contained in the intersection support of $f$, so in particular has compact support. (It's important to remember here that "compact support" means the set of non-zero points is contained in a compact set, not that it is compact.) Furthermore the support of $f+g$ is contained in the union of the supports of $f$ and $g$, so with a nod to the fact that products and sums of continuous functions are continuous, $(b)$ does define an ideal.

Finally take something like $\frac{1}{1+x^2}$ in the set of functions defined by $(c)$. There's an apparent continuous function we can multiply this by to get $1$, yes? And $1$ does not vanish towards $\infty$.