Injectivity of a linear operator

[D_Miller]

Technically this isn't homework, but just something I saw another user state without proof in a very different thread. I believe, however, that it is specific enough to pass as a "homework question" so I thought I'd pretend that it was and post it here, because I'm getting a bit frustrated with it:

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.

Part (a) is rather trivial, but part (b) bugs me. I have never seen this result before, and I cannot seem to prove it. My original idea was to use the open mapping theorem in relation to the kernel of the operator, but I couldn't make it work. I still think the open mapping theorem or a similar result is the right way forward, but I would very much appreciate it if someone could write out a proof.

Oh, and if it could be of help to anyone, the injectivity of M_f follows when the set of zeroes of f has measure 0.

 

[arkajad]

I don't understand. If f(x)=0 for all x in [0,1] then M_f=0, so how can it be one-to-one?

 

[arkajad]

One quick observation is (that you probably know about) is that if the range is closed then the inverse operator (restricted to the range) is bounded, therefore f must never vanish. Conversely, if f never vanish, then the inverse exist on the whole space, thus, in particular, the range is closed. I hope the above is correct and that you will be able to fill the holes.

 

[micromass]

Hmm, I just accepted this fact in the other thread without thinking about it. But thinking about it, it seems to be less obvious. But what if you take f=1 (the constant 1-function). Then certainly this is one-to-one and has a closed range...

 

[arkajad]

So, what's wrong with that?