Proving that a certian subset of L^1([0,1]) is closed

[benorin]

The problem: Let M be the set of all $$f\in L^1 \left( \left[ 0,1\right] \right)$$ relative to the Lebesgue measure, such that

$$\int_{t=0}^{1}f(t)dt = 1$$.

Show that M is a closed convex subset of $$L^1 \left( \left[ 0,1\right] \right)$$ which contains infinitely many elements of minimal norm.

What I've got: Define the linear functional $$\Lambda f = \int_{t=0}^{1}f(t)dt$$

Convexity is easy, for any f,g in M and s in (0,1) put $$h_s(t)=sf(t)+(1-s)h(t)$$ so that

$$\Lambda h_s = \Lambda \left( sf+(1-s)h\right) = s\Lambda f + (1-s)\Lambda g = s + (1-s) = 1\Rightarrow \int_{t=0}^{1}h_s(t)dt = 1\Rightarrow h_s\in M$$

by which it is understood that M is convex.

That M is closed is my first trouble. Let $$\left\{ f_k\right\} \rightarrow f$$ be a sequence of vectors such that $$f_k\in M,\forall k\in\mathbb{N}$$. Then we know that

$$\forall \epsilon >0, \exists N\in\mathbb{N}\mbox{ such that }k\geq N\Rightarrow \| f_k - f\| = \Lambda \left( |f_k - f|\right) < \epsilon$$

How do I prove that $$f\in M$$ ? Can I show that $$\Lambda$$ is a bounded linear map (actually functional) from $$L^1 \left( \left[ 0,1\right] \right)$$ into $$\mathbb{C}$$?

 

[benorin]

Think I got the closed part:

$$|1 - \Lambda f |=|\Lambda f_k - \Lambda f |=|\Lambda \left( f_k - f\right) |\leq \Lambda \left( |f_k - f|\right) < \epsilon\Rightarrow 1-\epsilon < \Lambda f < 1+ \epsilon$$

anbd since $$\epsilon$$ is arbitrary, $$\Lambda f =1\Rightarrow f\in M,$$ right?

 

[matt grime]

I think you're making it more complicated than it is. I mean, what is the topology on the space? Surely it is then trivial that the inverse image of a the point 1 under the (continuous, from the topology you put on the space!) functional f--> integral(f) is closed.