# Thread: Closed subset of L2

1. ## Closed subset of L2

Hey, I'm just wondering how to show the following:

Let K be the subset of $L^2([0,1])$ consisting of the equivalence classes of non-negative measurable functions $f$ with $\int_0^1 f(t) \mathrm{d}t = 1$. Show that $K$ is closed in $L^2([0,1])$.

2. Originally Posted by Giraffro
Hey, I'm just wondering how to show the following:

Let K be the subset of $L^2([0,1])$ consisting of the equivalence classes of non-negative measurable functions $f$ with $\int_0^1 f(t) \mathrm{d}t = 1$. Show that $K$ is closed in $L^2([0,1])$.
Isn't $\varphi:L^2([0,1])\to\mathbb{R}:f\mapsto \int_{[0,1]}f\text{ }dx$ continuous where $\mathbb{R}$ has the usual topology? (I'm asking) and if so isn't $K=\varphi^{-1}(\{1\})$?

3. Originally Posted by Drexel28
Isn't $\varphi:L^2([0,1])\to\mathbb{R}:f\mapsto \int_{[0,1]}f\text{ }dx$ continuous where $\mathbb{R}$ has the usual topology? (I'm asking) and if so isn't $K=\varphi^{-1}(\{1\})$?
I think the best you can do is say that $\varphi : L^1([0,1]) \to \mathbb{C}, f \mapsto \int_{[0,1]} f$ is continuous. I've also got $\forall f \in L^2([0,1]), \|f\|_{L^1([0,1])} \leq \frac{1 + \|f\|_{L^2([0,1])}^2}{2}$. Also, if $X$ is the subset of $L^2([0,1])$ consisting of the equivalence classes of non-negative measurable functions, then $K = \varphi|_X^{-1}(\{1\})$.

Isn't there a result along the lines of convergence in $L^1$ implies the existence of a subsequence that converges pointwise almost everywhere? I can't seem to find it.

4. Originally Posted by Drexel28
Isn't $\varphi:L^2([0,1])\to\mathbb{R}:f\mapsto \int_{[0,1]}f\text{ }dx$ continuous where $\mathbb{R}$ has the usual topology? (I'm asking) and if so isn't $K=\varphi^{-1}(\{1\})$?
This works fine, but you would have to use one of the following results:

1) If $\mu (X) < \infty$ then $L^p (X) \subset \L^q (X)$ for all $1\leq q \leq p$

2) If $f\in L^p (X)$ , $g \in L^q (X)$ where $\frac{1}{p} +\frac{1}{q} =1$ then $fg\in L^1 (X)$ and the for fixed $f$, the map $g\mapsto \int_X fg$is continous (Riesz representation theorem).

Another approach that works for arbitrary $X$ and $p$ is taking $A_m:= \{ f\in L^p (X) : f\geq 0 \, \ f(x)\leq m \ \ \mbox{and} \ \int_X f <\infty \}$ and using the monotone convergence theorem on $f_m = f|_{A_m}$.

5. Originally Posted by Giraffro
I think the best you can do is say that $\varphi : L^1([0,1]) \to \mathbb{C}, f \mapsto \int_{[0,1]} f$ is continuous.
The argument still applies.

Originally Posted by Jose27
This works fine, but you would have to use one of the following results:

1) If $\mu (X) < \infty$ then $L^p (X) \subset \L^q (X)$ for all $1\leq q \leq p$

2) If $f\in L^p (X)$ , $g \in L^q (X)$ where $\frac{1}{p} +\frac{1}{q} =1$ then $fg\in L^1 (X)$ and the for fixed $f$, the map $g\mapsto \int_X fg$is continous (Riesz representation theorem).

Another approach that works for arbitrary $X$ and $p$ is taking $A_m:= \{ f\in L^p (X) : f\geq 0 \, \ f(x)\leq m \ \ \mbox{and} \ \int_X f <\infty \}$ and using the monotone convergence theorem on $f_m = f|_{A_m}$.
Hmm, I see.

6. Heh, a slight typo on my post $A_m \subset X$ should be the set that satisfies $f\geq 0$ and $f\leq m$ on $A_m$ otherwise it makes no sense.