Recent content by stradlater
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Graduate What is a free sequence in a Hilbert space?
Thanks, linearly independent is exactly what it meant :) So it's a sequence of linearly independent vectors that span the entire space. It makes sense because the author states that this sequence is not necessarily orthonormal.- stradlater
- Post #5
- Forum: Calculus
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Graduate What is a free sequence in a Hilbert space?
Yes, it's the book Numerical Analysis by Roger Temam. Translated from french.- stradlater
- Post #3
- Forum: Calculus
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Graduate What is a free sequence in a Hilbert space?
Hi! I'm reading a book on the finite element method and the author mentions a free and total sequence in a hilbert space. I've been searching the internet, but I just can't find the definition of a free sequence. Does anybody know what it is? Thanks in advance- stradlater
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- Sequence
- Replies: 4
- Forum: Calculus
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Graduate Smooth and L^2 on R^n. Will it be bounded?
To make things clearer, consider this example(I think this is the function g_edgar mentioned). Let \psi(x) be a smooth non-negative function such that its support is contained in the interval (-1,1) and such that \int_R\psi(x)dx=1. Now consider the series \sum_{k=1}^\infty k\psi ((x...- stradlater
- Post #5
- Forum: Calculus
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Graduate Smooth and L^2 on R^n. Will it be bounded?
Thanks for the quick answers. With smooth I just meant that all the derivatives exist and are continuous. So with smooth I meant to disallow for the above mentioned example regarding sets of measure zero. And yes, the question was about one single function, but I think the sequence of bumps...- stradlater
- Post #4
- Forum: Calculus
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Graduate Smooth and L^2 on R^n. Will it be bounded?
smooth and L^2 on R^n. Will it be bounded?? Hello, If a function, say u, is smooth and L^2 on R^n. Will it be bounded?? In the case of n=1 I would say that it obviously is so. Because if it were unbounded then it wouldn't be L^2. But in the case of n=2 (or higher). I can imagine a...- stradlater
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- Bounded Smooth
- Replies: 4
- Forum: Calculus