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Math Help - Another Convolution Problem

  1. #1
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    Another Convolution Problem

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

    OK, so I proved in a previous exercise that that if f is locally integrable and g is infinitely smooth with compact support, that the convolution is of class C^{\infty}. With that, the hint given is to consider a convolution of two functions: f which is the characteristic function on some interval (which is locally integrable) and g\in C^{\infty}_c(\mathbb{R}). So we know the convolution would be of class C^{\infty} which would take care of the smoothness aspect. I'm just not seeing how to produce 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 g (if there is no standard for you then just take a non-negative one with support in [-1,1]), and define  f(x)=\frac{\int_{-\infty}^x g(y)dy}{\int_{\mathbb{R}}g(y)dy}, now just put h(x)=f(x+T+1)f(T+1-x). Now it's standard to show that 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 J which properly contains [-T,T], now let c be the distance from [-T,T] to the complement of J, f the characteristic function of J and take your g such that its support is contained in, say, [-c/2,c/2]. Now it's easy to prove this functions satisfies what is asked (If you know that \mbox{supp} f*g \subset \mbox{supp}f + \mbox{supp} g).
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