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Thread: Another Convolution Problem

  1. #1
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    Another Convolution Problem

    Show that for every $\displaystyle T>0$, there is a smooth function with compact support $\displaystyle \chi : \mathbb{R}\longrightarrow[0,1]$ such that $\displaystyle \chi=1$ on $\displaystyle [-T, T]$.

    OK, so I proved in a previous exercise that that if $\displaystyle f$ is locally integrable and $\displaystyle g$ is infinitely smooth with compact support, that the convolution is of class $\displaystyle C^{\infty}$. With that, the hint given is to consider a convolution of two functions: $\displaystyle f$ which is the characteristic function on some interval (which is locally integrable) and $\displaystyle g\in C^{\infty}_c(\mathbb{R})$. So we know the convolution would be of class $\displaystyle C^{\infty}$ which would take care of the smoothness aspect. I'm just not seeing how to produce $\displaystyle g$....
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  2. #2
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    In the real line this problem has a nice, somewhat explicit solution: Take, for example, the standard smooth function with compact support $\displaystyle g$ (if there is no standard for you then just take a non-negative one with support in $\displaystyle [-1,1]$), and define $\displaystyle f(x)=\frac{\int_{-\infty}^x g(y)dy}{\int_{\mathbb{R}}g(y)dy}$, now just put $\displaystyle h(x)=f(x+T+1)f(T+1-x)$. Now it's standard to show that $\displaystyle h$ satisfies what is asked.

    On the other hand, the approach you're hinted at works to construct so called "cut-off" functions over any open bounded subset, even in higher dimensions. To prove it like this, take an interval $\displaystyle J$ which properly contains $\displaystyle [-T,T]$, now let $\displaystyle c$ be the distance from $\displaystyle [-T,T]$ to the complement of $\displaystyle J$, $\displaystyle f$ the characteristic function of $\displaystyle J$ and take your $\displaystyle g$ such that its support is contained in, say, $\displaystyle [-c/2,c/2]$. Now it's easy to prove this functions satisfies what is asked (If you know that $\displaystyle \mbox{supp} f*g \subset \mbox{supp}f + \mbox{supp} g$).
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