If $ f \in L^2 ( \Bbb R^n)$ , then $\lim_{x_k \to \infty} f(x_1 , \cdots , x_k , \cdots , x_n ) = 0$ ? If this is true, then how can I prove this?
If $ f \in L^2 ( \Bbb R^n)$ , then $\lim_{x_k \to \infty} f(x_1 , \cdots , x_k , \cdots , x_n ) = 0$ ?
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real-analysis
functional-analysis
1 Answers
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It's not true necessarily... even in one dimension you can let $f(x) = \sum_{m = 1}^{\infty}\chi_{[m, m + {1 \over m^2}]}(x)$. This is in $L^2$ but $\lim_{x \rightarrow \infty}f(x)$ doesn't exist. In higher dimensions you can use $f(x_k)\chi_{[0,1]^{n-1}}(x')$, where $x'$ are the $x$ variables other than $x_k$.
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3I usually encounter physics students who think that square-integrable functions should vanish at infinity. This is because of their physical intuition that a particle is very unlikely to be found in distant regions of space, so that means the particle wavefunction should vanish in these regions. However, mathematically, this is not true, as nicely shown above. – 2012-10-15