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Suppose we have the following lemma:

Lemma If $E_0 \hookrightarrow E$, and $E_0$ is a closed subspace then $E/E_0$ is a normed space and for $[x] \in E/E_0$ its norm is given by $||[x]|| = \text{inf}_{y \in E_0} ||x-y||$.

What is the intuition of this norm besides that it "works"? Does it just measure how close $x$ and $y$ are to being equivalent? Also does one usually try to introduce some norm on a space $E$ but finds out that it doesn't work? Thus one introduces it as a seminorm and then as a norm on $E/E_0$? In other words, is a seminorm on $E$ just a means to get a norm on $E/E_0$? Why can't one just introduce a norm on $E/E_0$ directly?

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up vote 9 down vote accepted

Your question seems to be superimposing several concepts/issues at once, and perhaps it will help to separate them.

First, if $E$ is a semi-normed space and $E_0$ is a subspace then we define the semi-norm on $E/E_0$ according to the formula you wrote down. If you look at what's written, the coset $[x]$ is fixed, while $y$ is varying, so there is no particular $y$ of which it makes sense to ask "how close $x$ and $y$ are to being equivalent". Rather, you are measuring the distance of the point $x$ from the subspace $E_0$. Note that this depends only on $[x]$ (i.e. if we translate $x$ by an element of $E_0$, the distance to $E_0$ doesn't change). This is the most natural semi-norm to place on the quotient $E/E_0$: the distance of the coset $[x]$ from the $0$ in $E/E_0$ is measured by considering the distance of the representative $x$ from the $0$ coset thought of as a subset of $E$, i.e. the distance of $x$ from $E_0$. If $E_0$ is in fact closed in $E$ (with respect to the topology defined by the semi-norm on $E$), then one sees that the semi-norm on $E/E_0$ is actually a norm (and conversely).

Your second question (which got tangled up with your first, but is really a separate question, I think) is why do we sometimes consider semi-norms rather than norms. There are some situations in which $E$ naturally comes with a seminorm which is not a norm. We can then consider the subspace $E_0$ consisting of all elements of semi-norm equal to $0$, which will be a closed subspace (in the topology defined by the semi-norm). (In fact, $E_0$ is precisely the closure of the point $0$.) Now we can apply the above discussion to this particular choice of $E_0$, and so we get a norm on the quotient $E/E_0$.

This procedure is applied for example to construct the $L^p$ spaces in measure theory. If we define $E$ to the be the space of all measurable real-valued functions on some measure space $X$ for which $\int |f|^p d\mu$ is finite (for some $p \geq 1$), then $(\int |f|^p d\mu)^{1/p}$ is a semi-norm on $E$. One finds that $E_0$ is precisely the subspace of $E$ consisting of functions which vanish a.e., and the $L^p(X)$ is defined to be the quotient $E/E_0$.

So the reason for not introducing $E/E_0$ directly in this case is that one first has to talk about functions, and then the elements of $E/E_0$ are certain equivalence classes of functions which it's not really possible to introduce directly, without first going via the functions themselves and then introducing the equivalence relation. And when we write down the "$L^p$-norm" on functions, which is what we naturally can write down, it turns out not be a norm, but just a semi-norm. Passing to $E/E_0$ then gives us an actual normed space, rather than just a semi-normed one.

As for why we pass from $E$ to $E/E_0$ at all, this is mainly convenience. From the point of view of analysis, there is essentially no difference to working in $E$ and $E/E_0$; if any two functions in $E$ lie in the same $E_0$-coset (i.e. they coincide a.e.) then you can't really tell the difference between them analytically. So it's just convenient to identify them, and to pass to their common coset in $E/E_0$. (A general priciple is that it's simpler to work in Hausdorff spaces, like $E/E_0$, than in non-Hausdorff ones, like $E$ itself, and so when nothing is lost by passing from $E$ to $E/E_0$, it's easiest just to do so.)

Finally, are seminorms purely an intermediate step on the way to norms, as in the example of $L^p$-spaces? The answer is no. The reason is that in some situations one has whole families of semi-norms on a space (which are not necessarily themselves norms), which define an interesting or important topology.

E.g. consider the space $E$ of all continuous functions on $\mathbb R^n$. This doesn't have a natural norm: the obvious norm to consider on continuous functions is the sup norm, but since $\mathbb R^n$ is not compact, a continuous function need not be bounded, and so the sup norm is not defined on arbitrary continuous functions.

But, if $K$ is a compact subset of $\mathbb R^n$, then we can define a semi-norm $|| \quad ||_K$ on $E$ as $|| f ||_K =$ the sup of $f$ on $K$. Note that this is not a norm: there can be continuous functions on $\mathbb R^n$ which are not identically zero, but vanish at every point of the particular compact set $K$. But we can use this whole family of semi-norms to define a topology: we define a base of open sets by defining, for each $K$ and each $\epsilon >0$, a n.h. $B_{K,\epsilon}(f)$ of the function $f$ to be the set of $g$ such that $|| f - g||_K < \epsilon.$ This does form a basis of neighbourhoods of $f$, because given $K,\epsilon$ and $K',\epsilon',$ if we choose $K'' = K \cup K'$ and $\epsilon''$ to be the min of $\epsilon$ and $\epsilon'$, then $$B_{K'',\epsilon''}(f) \subset B_{K,\epsilon}(f) \cap B_{K',\epsilon'}(f).$$ This is a Hausdorff topology on $E$ (basically because a non-zero function must be non-zero when restricted to some compact set, so not all these semi-norms can vanish on a non-zero function), making $E$ a so-called Frechet space.

So here the semi-norms are appearing in their own right, not just as an intermediate step to getting a normed space.

So the passage from semi-norms to norms that you ask about is not really about getting rid of semi-norms and replacing them by norms at all: it is rather about getting rid of non-Hausdorff spaces and replacing them by Hausdorff ones. It just happens that if your topology is defined by a single semi-norm that is not a norm, then it will be non-Hausdorff, and when you make it Hausdorff (by taking the quotient by the closure of $0$) the resulting space has its topology defined by a norm. But in general, there are lots of Hausdorff spaces whose topology is defined not by a norm, but rather by a family of semi-norms. (In applications, these spaces appear most often in the foundational aspects of distribution theory.)

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Thank you for this wonderful explanation. – PEV Jan 6 '11 at 3:58
@Trevor: Dear Trevor, You're welcome. Regards, – Matt E Jan 6 '11 at 4:15

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