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Let $M$ be the set of all $f\in L^1([0,1])$, relative to Lebesgue measure, such that $$\int_0^1f(t)\,dt=1.$$ Show that $M$ is a closed convex subset of $L^1([0,1])$ which contains infinitely many elements of minimal norm.

For convexity - Let $f,g \in M \subset L^1([0,1])$ and let $\alpha\geq0$ and $\beta\leq1$ such that $\alpha+\beta=1$. So, $$\int_0^1(\alpha f+\beta g)(t)dt=\alpha+\beta=1$$ so M is convex.

For closure it's sufficient to show that $\int_0^1f(t)\,dt$ is continuous since if f is continuous, then the preimage of a closed set is itself a closed set. note that under this function, the preimage of the set {1} is itself $M$ and {1} is closed. To see this, let {$f_n$} $\subset M$ and suppose $||f_n-f||_1$ goes to zero. Then, $$\int_0^1f_n(t)dt-\int_0^1f(t)dt=\int_0^1f_n(t)-f(t)dt\leq\int_0^1|f_n(t)-f(t)|dt=||f_n-f||_1$$ which goes to zero as $n$ goes to infinity. Thus continuity is established and so M is a closed convex subset.

Also I really don't have a good idea of how to start the minimal norm argument. I do know that there is at least one element of minimal norm once M is closed and convex, but infinitely many is eluding me... help please!

sjf2468
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    Your "closed" argument is actually a convexity argument. You need to separately argue closedness. This is quick: since ${ 1 }$ is closed, it is enough to show that $f \mapsto \int_0^1 f(t) dt$ is continuous, since $M$ is the preimage of ${ 1 }$ under this function. But this is just the triangle inequality. The minimal norm part is harder, I think. – Ian Mar 31 '15 at 00:17
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    For the minimal norm part, note that the minimal norm is at least $1$ by the triangle inequality, and actually is $1$ since it is attained by the constant function $1$. So positive functions in $M$ have minimal norm. Can you make infinitely many of these? – Ian Mar 31 '15 at 00:25
  • Yes, that is what I meant. I apololgize for any confusion, and I deleted that comment. – shalop Mar 31 '15 at 13:15
  • Thank Ian. I edited my post with some additional info. Could you take a look and see if it is sufficient? Also, I still am unclear about the whole minimal norm leg of the proof...could you give a proof?Hints? – sjf2468 Apr 02 '15 at 00:03
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    You do not get that images of closed sets are closed, you get that preimages of closed sets are closed. – Ian Apr 02 '15 at 00:05
  • I see my error. Thanks again Ian. I am still in the dark on the last portion... I get that $1=\int_0^1f(t)dt \leq \int_0^1|f(t)|dt=||f||_1$ but don't have a foot-hole to finish...could you elaborate please? – sjf2468 Apr 02 '15 at 21:11
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    I've got it now. Define $f(t)=nt^_n-1$ and we have the result. yes? – sjf2468 Apr 02 '15 at 21:37

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