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

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

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

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

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

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

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

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

5. Originally Posted by Giraffro
I think the best you can do is say that $\displaystyle \varphi : L^1([0,1]) \to \mathbb{C}, f \mapsto \int_{[0,1]} f$ 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 $\displaystyle \mu (X) < \infty$ then $\displaystyle L^p (X) \subset \L^q (X)$ for all $\displaystyle 1\leq q \leq p$

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

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

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