A simple example
Let $f_n = n \cdot \mathbb{1}_{[0, 1/n)}$
the function converges pointwise to 0. Can I also say that the support of f is shrinking or is it best to just keep the phrasing: the function converges pointwise to 0?
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The sequence doesn't converge to $0$ at $0$ though. – Arthur Aug 05 '19 at 07:19
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2@Arthur : though $g_n = n \cdot \mathbb{1}_{(0, 1/n)}$ would – Henry Aug 05 '19 at 07:30
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that's fine since it's pointwise and not uniform – user29418 Aug 05 '19 at 11:06
2 Answers
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In this example the support is shrinking indeed. However, the information contained in the statement "the support is shrinking" is not sufficient in itself to guarantee pointwise convergence.
uniquesolution
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what is the correct statement btw? is "shrinking support" correct? also, what does vanishing support mean? I might be getting it mixed up a bit here. – user29418 Aug 05 '19 at 11:08
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No. If $f_n\colon\mathbb R\longrightarrow\mathbb R$ is defined by $f_n(x)=\frac1n$, then $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function, but the support of each $f_n$ is equal to $\mathbb R$,
José Carlos Santos
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That seems to ignore the $\mathbb{1}_{[0, 1/n)}$ term. The support of the original function $f_n$ looks to me like $[0,\frac1n]$ – Henry Aug 05 '19 at 07:19
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That is supposed to be just an example. The actual question is “does pointwise convergence mean that the support is shrinking?” and my answer explains that the answer to that question is negative. – José Carlos Santos Aug 05 '19 at 07:26
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OK - my confusion was the use of the same $f_n$ notation for a different example. My guess is that at some stage the question will motivated by probability density functions with $\int f_n(x) , dx =1$ – Henry Aug 05 '19 at 07:27
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This answer was also really good, thank you. It clarifies it for me. – user29418 Aug 05 '19 at 11:03