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Problem of the day:

Show that \((\mathcal{C}^1([0,1]),d_p)\) is a metric space for \(p=2\) given that \[d_p(f,g)=\left(\int_0^1|f(x)-g(x)|^pdx\right)^{\frac{1}{p}}.\] (Hint: the Cauchy-Schawrz inequality may prove itself useful).

Bonus problem:

Show that \((\mathcal{C}^1([0,1]),d_p)\) is a metric space \(\forall p \in \mathbb{N}-\{0\}\).

As usual, a brief explanation of the concepts dealt with in the problem is found below.


Supplementary Theory

Given a non-empty set \(X\), we say that the pair \((X,d)\) is a metric space if, and only if, \(d:X\times X\to\mathbb{R}\) is a function such that:

  1. \(d(x,y)\geq 0 \ \forall x,y\in X\)
  2. \(d(x,y)=0\Leftrightarrow x=y \ \forall x,y\in X\)
  3. \(d(x,y)=d(y,x) \ \forall x,y\in X\)
  4. \(d(x,y)\leq d(x,z)+d(z,y) \ \forall x,y,z\in X\)

Usually, we call the last property by the name of "triangle inequality" and we say \(d\) is a metric or a distance function. In fact, these four properties are unnecessary, since the first and the third one are consequences of the other two (it's actually not hard to see if you play with them for a while and it's insanely useful: it makes showing that a pair is, in fact, a metric space much faster).

Let us define the function \(d_{0,1}:X^2\to\{0,1\}\) such that \(d_{0,1}(x,y)=0\) for \(x=y\) and \(d_{0,1}(x,y)=1\) for \(x\not=y\). It's quite simple to see that \(d_{0,1}\) turns every pair \((X,d_{0,1})\) into a metric space.

This is more than enough theory for the problem.