1

In his book "Real and Complex Analysis", chapter "Elementary Hilbert Space Theory", Walter Rudin introduces the following Lemma to prove Bessel's inequality:

Given a complete metric space $X$, if $f:X \longrightarrow Y$ is a continuous function defining an isometry on a dense subset $X_0\subseteq X$, and if $f(X_0)$ is dense in $Y$, than $f$ is an isometry on $X$ and $f$ is surjective.

The proof details why $f$ is surjective, but only states that $f$ being an isometry is an immediate consequence of $f$ being continuous and $X_0$ being dense in $X$.

Why is that so? I do not see the immediate consequence.

PVAL-inactive
  • 8,551
Kuifje
  • 9,584
  • 4
    The map $D: X\times X \to \Bbb R$ with $D(x_1,x_2)=d_Y(f(x_1),f(x_2))-d_X(x_1,x_2)$ is continuous and you know what it is equal to over a dense subset of $X\times X$. – PVAL-inactive Sep 22 '15 at 21:45
  • Ok, so the idea is to say that it is equal to zero on $X_0 \times X_0$ because $f$ is an isometry on $X_0$, and since it is continuous (why so?) it has to remain equal to zero for points not in $X_0$? How do we use both assumptions ($f$ continuous and $X_0$ dense)? – Kuifje Sep 22 '15 at 22:08
  • If two continuous functions are equal over a dense set, they are equal over the entire space (prove this). $D$ can be shown to be a composition of continuous functions (the map $X\times X \to \Bbb R$ where $(x_1,x_2) \mapsto d_X(x_1,x_2)$ is pretty easy to show continuity). – PVAL-inactive Sep 22 '15 at 23:32
  • 1
    Let me sum up:

    Let $D$ be the continuous map as you defined it. And consider the map $\hat{D}$ defined on the same space as $D$, with $\hat{D}(x_1,x_2)=0 ; \forall x_1,x_2$.

    $D$ and $\hat{D}$ agree on the subspace $X_0\times X_0$, and therefore they are equal on $X\times X$(a proof is given for example here: http://math.stackexchange.com/questions/202325/continuous-functions-between-metric-spaces-are-equal-if-they-are-equal-on-a-dens).

    It follows that $d_Y(f(x_1),f(x_2))-d_X(x_1,x_2)=0$ for all $x_1,x_2$, which proves that $f$ is an isometry on $X$.

    Does this seem correct?

     – Kuifje Sep 23 '15 at 13:29

1 Answers1

1

Just to sum up the comments above, which already completely answered the question:

the continuous maps $d_X:X\times X\to\Bbb R$ and $$(x,y)\mapsto d_Y(f(x),f(y))$$ coincide on the dense subspace $X_0\times X_0,$ hence they coincide everywhere.

Anne Bauval
  • 34,650