# Linear functional clarification (from rudin)

1. Aug 28, 2009

### redrzewski

In Rudin's Functional Analysis (in theorem 3.4), he says:

"every nonconstant linear functional on X is an open mapping". X is topological vector space.

This seems like a strengthening of the open mapping theorem, which requires X to be an F-Space, and that the linear functional to be continuous.

Indeed, it seems like a discontinuous linear functional from R to R as described in Gelbaum for instance isn't open.

What am I missing?
thanks

2. Aug 28, 2009

### g_edgar

There is no discontinuous linear functional R to R. Linear is in the sense of real scalars.

3. Aug 28, 2009

### redrzewski

Are you sure? "Counterexamples in Analysis" by Gelbaum p.33 gives an example.

4. Aug 28, 2009

### g_edgar

p. 33 ... compare it carefully to what you said before.

5. Aug 28, 2009

### redrzewski

This isn't homework. Can you be more direct?

As far as I see, the function defined by Gelbaum is a discontinuous linear functional R to R.

6. Aug 28, 2009

### redrzewski

I think the problem is that Gelbaum's construction is using a different definition of linear. It seems Gelbaum's "linear" doesn't also imply f(a*x) = a*f(x) for scalar a. So that resolves some of my confusion. Does this sound accurate?

That still leaves my original question, though:

Is "every nonconstant linear functional on X is an open mapping"? Even assuming that the functional is continuous, do we really not need the stronger condition that X is an F-space?

7. Aug 29, 2009

### g_edgar

We do not need $F$-space and we do not need continuous. A linear functional $f$ on a TVS $X$ can be discontinuous. But the point of the "hints" I gave before is that when you restrict $f$ to a line (say a line $L = \{tu : t \in \mathbb{R}\}$ through the origin and the point $u$), that restriction is continuous, in fact it is of the form $f(tu) = tc, t \in \mathbb{R}$, for some constant $c$. And (more to the point) it maps open intervals in $L$ to open intervals in $\mathbb{R}$.

We do have to assume $f$ is not the zero linear functional!

We claim $f$ is an open map. Let $U \subset X$ be an open set. We must show the image $f(U)$ is open in $\mathbb{R}$. Let $u \in U$ . We claim that $f(U)$ contains an interval centered at $f(u)$. This is because of the restriction property mentioned above. OK?

(Strictly speaking, we still have to adjust something when $f(u) = 0$.)

8. Aug 29, 2009

### wofsy

Your post reminds me of the example of a linear functional from R to R when R is viewed as a vector space over the rationals.

It is easy to construct a discontinuous one from a Q basis.

Problem: Show that such the graph such a discontinuous linear functional is dense in the plane.

9. Aug 30, 2009

### redrzewski

Thanks, g_edgar. That clears things up.

wofsy: that's the same construction that Gelbaum is citing. Am I correct that that construction doesn't have the property that: f(a*x) = a*f(x) for scalar a?

10. Aug 30, 2009

### wofsy

if a is a rational number, then f(a*x) = a*f(x).

The problem has more parts to it. Show that if f is measurable then it is continuous. If f is continuous then it is just multiplication by a scalar.
What does that tell you about the measurability of a basis for the reals over the rationals?

Last edited: Aug 30, 2009
11. Oct 28, 2010

### tinchote

I don't see the "adjust something" of g_edgar very clearly (for the case f(u)=0).

None of that is required to complete Rudin's argument, though. In his case, what one can do is simply let gamma=sup Gamma(A). All one needs to prove is that gamma is not in the image of A. Assume that it is in the image of A; that is, gamma=Gamma(a) for some a in A. Then, as A is open, one can find a balanced neighbourhood U of 0 with a+U in A. If u in U, then gamma >= Gamma(a+u)=gamma+Gamma(u), so we conclude that Gamma(u)<=0. But, as U is balanced, this argument should also work for -u, and that gives us a contradiction. The conclusion is that gamma is not in the image of A, and so Gamma(a)<gamma for all a in A.