# Continuous Functions, Closed Sets

1. Apr 8, 2008

### PingPong

1. The problem statement, all variables and given/known data

A mapping f from a metric space X to another metric space Y is continuous if and only if $f^{-1}(V)$ is closed (open) for every closed (open) V in Y.

Use this and the metric space (X,d), where X=C[0,1] (continuous functions on the interval [0,1]) with the metric $d(f,g)=\sup _{x\in[0,1]}|f(x)-g(x)|$ to show that the set
$$A=\left\{f\in C[0,1]:\int_0^1 tf^2(t)dt\ge 3\right\}$$

is closed.

2. Relevant equations
None other than the theorem given.

3. The attempt at a solution

My first step is to define a function from C[0,1] to R by choosing the easiest function, $F(f)=\int_0^1 tf^2(t)dt$. If I can show that this is continuous, then it follows that $\left\{f\in C[0,1]:\int_0^1 t f^2(t)dt\ge 3\right\}$ is closed since $[3,\infty)$ is closed. That is, A is closed.

So, starting from the definition of a continuous function, I have:
$$|F(f)-F(g)|=\left|\int_0^1 t\left(f^2(t)-g^2(t)\right)\right|dt \le \int_0^1\left|t\left(f^2(t)-g^2(t)\right)\right|dt \le \int_0^1 |f^2(t)-g^2(t)|dt=\int_0^1 |f(t)-g(t)||f(t)+g(t)|dt$$
$$\le \int_0^1 d(f,g)|f(t)+g(t)|dt=d(f,g)\int_0^1 |f(t)+g(t)|dt<\epsilon$$.

My trouble is, I can't get rid of the last part of the integral that has |f+g| without restricting the domain of F. Any suggestions? Also, is my reasoning here fine? Thanks in advance!

2. Apr 8, 2008

### morphism

This may be easier to see using sequences. A useful thing to keep in mind is that convergence w.r.t. d is uniform convergence. Now, if f_n -> f uniformly, can we say that |f_n| -> |f| uniformly, too? How can this help us here? (Note: We also have that f_n+f -> 2f uniformly, since d(f_n+f,2f)=d(f_n,f)->0.)

3. Apr 9, 2008

### PingPong

Okay. Sorry if I seem slow here, but I'm trying to restate what you're saying so that I understand it.

Reinterpreting what I've written with a sequence instead of g, I have

$$|F(f)-F(f_n)|=\left|\int_0^1 t\left(f^2(t)-f_n^2(t)\right)\right|dt \le \int_0^1\left|t\left(f^2(t)-f_n^2(t)\right)\right|dt \le \int_0^1 |f^2(t)-f_n^2(t)|dt=\int_0^1 |f(t)-f_n(t)||f(t)+f_n(t)|dt$$
$$\le \int_0^1 d(f,f_n)|f(t)+f_n(t)|dt=d(f,f_n)\int_0^1 |f(t)+f_n(t)|dt<\epsilon$$

Now, $\int_0^1 |f(t)+f_n(t)| dt$ goes to a finite number, so I can pick $\delta=\frac{\epsilon}{\int_0^1 |2(f)| dt}$. Thus F is uniformly continuous (and hence continuous), and so I can apply the theorem. So A is closed.

This doesn't totally make sense to me, so if what I've done doesn't make sense then that's why.

Thanks again!

4. Apr 9, 2008

### morphism

Since the sequence of real numbers $\{\int_0^1 |f(t)+f_n(t)| dt\}_n$ converges, it's bounded by some M>0.

Hence, given $\epsilon > 0$, we can find an N such that if n>N then $d(f,f_n) < \epsilon/M$, and consequently, $|F(f) - F(f_n)| \leq d(f,f_n) \cdot M < \epsilon$. In other words, F(f) -> F(f_n).

5. Apr 9, 2008

### PingPong

Okay, that makes more sense now. I see where you're pulling things from. Thank you!