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I am struggling to follow a book on finite elements method, and I believed I had understood what is a $u \in L²(a, b)$ . In my words, $u$ is any function $u: (a, b) \rightarrow \mathbb{R}$, where $(a,b)$ is a range in $\mathbb{R}$, that behaves "nicely" (never escapes to infinity, neither have other oddities so that it satisfies $\int_a^bu(x)^2 dx \in \mathbb{R}$ ).

I can relate $u$ to a point in the infinite dimension Hilbert space $L²(a,b)$ by thinking that, the same way that $(x, y, z)$ is a point of $\mathbb{R}^3$, providing one coordinate to each dimension of the space, the continuous functions $u$ provides one value for each of the infinite real values between $a$ and $b$. Is that a correct intuition?

If so, what does it means to say $V^h \subset L^2(a,c)$, where $h$ is a length inversely proportional to the finite number of dimensions? What kind of function $v$ is both in $V^h$ and $L^2(a,c)$? How can be both a set of finite discrete coordinates and a continuous curve? At the very least, the domain of a function in $V^h$ should be discrete, so it can't be exactly the same in $L2(a,b)$, right?

lvella
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  • "Is this a correct intuition"? More or less. Be aware that there are pitfalls, if you reason like that: this train of thoughts might lead you to think that the set of functions $$\left{ f_y\colon (a, b) \to \mathbb{R},\ f_y(x)=\begin{cases} 0 & x\ne y \ 1 & x=y\end{cases}\right}$$ is a "basis" of some sort for $L^2(a, b)$, but unfortunately this is very far from being true. Indeed, any of the functions $f_y$ is such that $$\int_a^b f_y(x)^2,dx =0, $$ and so it is indistinguishable from the identically zero function in $L^2(a, b)$. One says that $f_y(x)=0$ almost everywhere. (HTH) – Giuseppe Negro Mar 12 '14 at 22:06
  • That thought had not occurred to me, mainly because the very notion of making a linear combination of a non-enumerable infinite set of functions is alien to me... it makes much more sense to me just to define the function itself. – lvella Mar 13 '14 at 02:07
  • That's good. Therefore you won't have any trouble in understanding $V^h$. It is just a set of linear combinations of a finite family of functions. In the finite element method (AFAIK), those are usually "tent functions", whose graph is sharply reverse-V shaped (don't take me too seriously - I'm no expert). Your set $V^h$ is just a family of linear combinations of those tent functions. I would like to remark that, since $V^h\subset L^2(a, b)$, any function contained in it is contained in $L^2(a, b)$ also, and so it is defined on the whole of $(a, b)$. – Giuseppe Negro Mar 13 '14 at 08:21

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