closed subset of a metric space

[r4nd0m]

This seems to be a very easy excercise, but I am completely stuck:
Prove that in C([0,1]) with the metric
\rho(f,g) = (\int_0^1|f(x)-g(x)|^2 dx)^{1/2}

a subset
A = \{f \in C([0,1]); \int_0^1 f(x) dx = 0\} is closed.

I tried to show that the complement of A is open - it could be easily done if the metric was \rho(f,g) = sup_{x \in [0,1]}|f(x)-g(x)| - but with the integral metric it's not that easy.

Am I missing something?
Thanks for any help.

[matt grime]

I would attempt to show that the map f--->int f(x)dx is continuous.

And you need / not \ in your closing tex tags.

[NateTG]

Consider if:
\left| \int_0^1 g(x)dx \right| = \epsilon
then you might be able to show that
N_{\epsilon}(g(x)) \cap A = \emptyset

[r4nd0m]

Well, I tried both, but the problem is that I still miss some kind of inequality that I could use.

I mean - if I want to show the continuity for example - I have to show that:
\forall \varepsilon > 0 \quad \exists \delta >0 \quad \forall g \in C([a,b]) : \rho(f,g)<\delta \quad |\int^1_0 f(x)-g(x) dx|< \varepsilon .

But what then?
|\int^1_0 f(x) - g(x) dx| \leq \int^1_0 |f(x) - g(x)| dx
but I miss some other inequality where I could compare it with \rho(f,g)

[matt grime]

Well, there is another inequality lying around. So use it That last inequality is also <=p(f,g).

[r4nd0m]

Of course :rolleyes: - a little modified Cauchy-Schwartz inequality is the key.
I hate algebraic tricks :smile:
Thanks for help

[matt grime]

It's definitely not an algebraic trick. It is an application of the Jensen inequality from analysis. You might know it from probability theory, since it just states that the variance of a random variable is positive, i.e.

E(X^2)>E(X)^2

where E is the expectation operator and X an r.v.

[kaushik200]

How did you prove it for the supremum case? If you can prove it for the supremum, this proof here is self-contained because the given metric is always less than the supremum.