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**1. Homework Statement**

A mapping f from a metric space X to another metric space Y is continuous if and only if [itex]f^{-1}(V)[/itex] 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 [itex]d(f,g)=\sup _{x\in[0,1]}|f(x)-g(x)|[/itex] to show that the set

[tex]A=\left\{f\in C[0,1]:\int_0^1 tf^2(t)dt\ge 3\right\}[/tex]

is closed.

**2. Homework 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, [itex]F(f)=\int_0^1 tf^2(t)dt[/itex]. If I can show that this is continuous, then it follows that [itex]\left\{f\in C[0,1]:\int_0^1 t f^2(t)dt\ge 3\right\}[/itex] is closed since [itex][3,\infty)[/itex] is closed. That is, A is closed.

So, starting from the definition of a continuous function, I have:

[tex]|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[/tex]

[tex]\le \int_0^1 d(f,g)|f(t)+g(t)|dt=d(f,g)\int_0^1 |f(t)+g(t)|dt<\epsilon[/tex].

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!