Re: linear operator as dynamical system
- From: William Elliot <marsh@xxxxxxxxxxxxxxxxxx>
- Date: Fri, 29 Feb 2008 20:02:22 -0800
Thank you Robert and Mariano for clarifying usage of 'operator' in the
context of vector or linear spaces. In topology 'map' is used in a
similar way, it can mean a function or a continuous function, depending
upon context and author
.Just what is a linear operator f:V -> V?
A function f:V -> V or f:V -> W for which
f(x + y) = f(x) + f(y)
f(ax) = af(x) ?
Yes.
Why is the expression 'operator' used instead of function?
Why not?
In any case, an operator is just a function and the difference
is only psychological and historical.
Does it compare with the expression 'functional' which indicates
a function or map from a function space to a function space?
Usually functional is used for linear maps from vector
spaces to *scalars*.
assuming V and W have the same scalars.
That's covered by the "over the same field F"
What makes a linear operator closed?
The statement was about the field, not the operators.
I was pulling from another problem which asked for an example of
a closed operator that mapped a closed set to a not closed set.
In the context of functional analysis, a linear operator
f : V --> W is closed if its graph
{ (v, f(v)) in V direct sum W : v in V }
is a closed subspace of V direct sum W.
That has nothing to do with the present context.
Perhaps this is a different sense of operator, yet it was an analysis
problem where linear spaces are in vogue and 'operator' could likely be
used for 'linear function'. Is that common usage with vector spaces,
that an operator is usually linear?
It is common when one is dealing with linear operators
to say simply `operator'. When one is dealing with
non-linear operators, one also says simply `operator'.
In general, one uses whatever is more convenient in the
context that one is in.
Since + is continuous
lim(x->r) f(x) = lim(x->0) f(r + x)
= lim(x->0) (f(r) + f(x))
= f(r) + lim(x->0) f(x)
So if a linear function is continuous at 0, it's continuous.
Indeed.
By the continuity of scalar multiplication, can I get away with
lim(x->0) f(x) = lim(a->0) f(ax) = lim(a->0) a.f(x) = 0 = f(0)
or
lim(x->0) f(x) = lim(a->0) f(ay) = lim(a->0) a.f(y) = 0 = f(0);
that linear functions are continuous at 0, hence continuous?
It is not true that
lim(x->0) f(x) = lim(a->0) f(ax)
Moreover, there are linear maps which are not continuous.
To provide an example, you need infinite dimensional vector
spaces (a linear map between finite dimensional vector spaces
*is* continuous). A very simple example is the following:
Let V be the vector space of all functions f : R --> R
which are bounded and differentiable at 0. Consider the
sup-norm on V, so that
|| f || = sup_{x \in R} |f(x)|
and put on V the corresponding topology. Then the map
f in V ---> f'(0) in R
is not continuous, yet it is linear.
If I took a Hamel base of reals over rationals, would that suffice for
making a counter example?
A counter example to what? If to the statement that linear maps
are continuous, you'd have to build a map from the Hamel basis
somehow and *then* we'd know if it worked or not...
-- m
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