Lp-Spaces: Proving (a) & (b) for Continuous f & Generalizing (a)

[Hjensen]

I'm taking a course on Banach spaces and our lecturer proved the following result:

Let f be a continuous, complex-valued function on [0, 1] and define a linear operator $$M_f$$ on $$L^2[0,1]$$ by $$(M_f g)(x)=f(x)g(x)$$. Then
(a) $$M_f$$ is bounded and $$||M_f||=||f||_{\infty}$$.
(b) Suppose that $$M_f$$ is one-to-one. Then the range of $$M_f$$ is closed if and only if $$f(x)=0$$ for all x∈[0,1]. In that case $$M_f$$ is one-to-one and
onto, and the inverse map is bounded.

I have two questions related to this result. First, is there any condition I can impose on f in order to make sure that the operator is always one-to-one, so (b) holds? I feel this would make the result nicer, but I can't think of anything that seems to work.

More importantly, the professor said (without proof - more as a passing remark) that you can generalize (a) to the case with f not necessarily continuous but in $$L^{\infty}[0,1]$$. Then he noted that the same isn't true for (b), but he didn't go into detail. Can anyone think of a concrete example as to why this generalization would not work?

I'd appreciate any help :).

 

[micromass]

First, I'd like to say that b is quite a negative result. It proves that $$M_f$$ almost never has a closed range and is almost never bijective.

As for your one-to-one question. I conjecture that $$M_f$$ is one-to-one iff the set of zeroes of f has measure 0. For example, if f never becomes zero, then $$M_f$$ is injective and thus doesn't have closed range.

b is not true for $$f\in L^\infty[0,1]$$. Consider the function

$$f(x)=\left\{\begin{array}{cc} 1 & x=0\\ 0 & x\neq 0\end{array}\right.$$

Then the function is not zero, but does have closed range.
A nice question is whether b holds if you replace "f=0" with "f=0 almost everywhere"?

 

[Hjensen]

Thanks a lot for your reply. I understand your arguing, but do you think a formal proof would require the continuity condition? I am curious as to whether this could be generalized to the case of discontinuous functions on $$L^{\infty}[0,1]$$ as well.

And you're absolutely right. It's not the most rewarding result I've ever seen either.

 

[micromass]

I am sorry, the function f I provided is not good, since f=0 a.e., thus f=0 in $$L^\infty[0,1]$$.

A better function would be

$$f(x)=\left\{\begin{array}{cc} 1 & x\leq 1/2\\ 0 & x> 1/2\end{array}\right.$$

That said, I think that the proof of b does depend on the continuity of f...

 

[Hjensen]

Oh absolutely, and it does. Perhaps I should've been more specific with with my question. The proof I was referring to was a formal proof of the idea that $$M_f$$ is one-to-one iff the set of zeroes of f has measure 0. Could this be done without the notion of continuity of f? I'm kinda guessing the answer is yes, but I'm not sure how one would do it.

 

[micromass]

Well, let N be the set where f vanished.
Assume that N has measure 0. Take a g such that $$M_f(g)=0$$. Then f(x)g(x)=0 for all x. Thus we have that

$$f(x)\neq 0~\Rightarrow~g(x)=0$$

So, if f only becomes 0 on a set of measure 0, this yields that g is only nonzero on a set of measure 0. Thus g=0 in $$L^2[0,1]$$.

On, the other hand, assume that N doesn't have measure 0. Take g the characteristic function of N, then g is not the zero function. But $$M_f(g)=0$$.

 

[Hjensen]

Thanks a lot for your time and effort. I'm pretty sure I get it.On a different, but somewhat related note, you know how the kernel of a continuous projection P=P² (or any other continuous linear operator) is closed? I seem to remember a very similar result for the range/image of a continuous projection? Meaning it's closed. Can you confirm this? I don't need a proof, it's just means to an end, and algebra isn't exactly my passion. :P

 

[micromass]

I think a linear operator is continuous iff it's kernel is closed.
Let f be continuous, then the inverse image of any closed set is closed. Thus $$f^{-1}(0)$$ is closed since {0} is closed.
The inverse implication is a lot less trivial (and I'm not 100% certain that it's true).

 

[Hjensen]

Yes, but is the range of a continuous projection not closed in general? Or maybe it has to be from a space into the same space? The kernel remark was just for perspective because I remembered that result.

 

[micromass]

Ah yes, I understand. Indeed, the range of a continuous projection operator is always closed...

 

[Hjensen]

That's what I thought. I thank you for all your patience and insight. I'm just a chemist trying to make it in the world of physics, so I salute people like you. :)

 

[Landau]

I think a linear operator is continuous iff it's kernel is closed.
Let f be continuous, then the inverse image of any closed set is closed. Thus $$f^{-1}(0)$$ is closed since {0} is closed.
The inverse implication is a lot less trivial (and I'm not 100% certain that it's true).

It is true for linear functionals on any topological vector space.

 

[D_Miller]

I wonder if somebody would care to present a proof of result (b) in Hjensen's original post. I have never seen that result before and I am probably missing something obvious but I cannot seem to make sense of it. It looks like one of those results that follows from the open mapping theorem or something similar, but I may be wrong.

 

[D_Miller]

Edit: False alarm. I thought I had figured it out, but I hadn't. I would still like to see a proof if Hjensen or anyone else could be bothered.