modern analysis: metric spaces and $\varepsilon$-neighborhoods

Question

Prove or disprove that $d(f,g) = ({\int_0^1 |f(x)-g(x)|^{2}dx})^{1/2}$, on $C[0,1]$ is a metric. If so, describe the $\varepsilon$-neighborhood.

Answer 1

Check the properties of a metric:

• $d(f,g) =0 $ if and only if $f=g$

• $d(f,g)=d(g,f)$ for all $f,g$

• $d(f,g) \geq 0$ for all $f,g$
• $d(f,g) \leq d(f,h) + d(h,g)$ for all $f,g,h$

Then, fix $f$ and see which $g$ are such that $d(f,g) \leq \epsilon$.

Answer 2

Do you know that for $f\in C[0,1]$ $$||f||_2=\left(\int_0^1|f(x)|^2dx\right)^{1/2}$$ is it's euclidean norm and then $$d(f,g)=||f-g||_2$$ is the metric associated to this norm?

Answer 3

Almost all properties are obvious, except probably the triangle inequality. This follows if we can show that $$\left(\int_0^1 |f+g|^2\right)^{1/2}\leqslant \left(\int_0^1 f^2\right)^{1/2}+\left(\int_0^1 g^2\right)^{1/2}$$

Upon squaring, we get $$\int_0^1 (f+g)^2\leqslant \int_0^1 f^2+2 \left(\int_0^1 g^2\right)^{1/2}\left(\int_0^1 f^2\right)^{1/2}+\int_0^1 g^2$$

And by the binomial theorem, we're reduced to showing $$\int_0^1 |fg|\leqslant \left(\int_0^1 g^2\right)^{1/2}\left(\int_0^1 f^2\right)^{1/2}$$

This is the known Cauchy-Schwarz inequality. There is a family of inequalities when $\dfrac 1q+\dfrac 1 p=1; p,q\geqslant 1$, namely $$\left(\int_0^1 |fg|^p\right) \leqslant \left(\int_0^1 |f|^p\right)^{1/p}\left(\int_0^1 |g|^q\right)^{1/q}$$ which gives Minkowski's inequality $$\left(\int_0^1 |f+g|^p\right)^{1/p}\leqslant \left(\int_0^1 |f|^p\right)^{1/p}+\left(\int_0^1 |g|^p\right)^{1/p}$$