# proof of Riesz representation theorem

Existence - If $f=0$ we can just take $u=0$ and thereby have $f(x)=0=\u27e8x,0\u27e9$ for all $x\in \mathscr{H}$.

Suppose now $f\ne 0$, i.e. $Kerf\ne \mathscr{H}$.

Recall that, since $f$ is continuous^{} (http://planetmath.org/ContinuousMap), $Kerf$ is a closed subspace of $\mathscr{H}$ (continuity of $f$ implies that ${f}^{-1}(0)$ is closed in $\mathscr{H}$). It then follows from the orthogonal decomposition theorem that

$$\mathscr{H}=Kerf\oplus {(Kerf)}^{\u27c2}$$ |

and as $Kerf\ne \mathscr{H}$ we can find $z\in {(Kerf)}^{\u27c2}$ such that $\parallel z\parallel =1$.

It follows easily from the linearity of $f$ that for every $x\in \mathscr{H}$ we have

$$f(x)z-f(z)x\in Kerf$$ |

and since $z\in {(Kerf)}^{\u27c2}$

$\mathrm{\hspace{1em}}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}0$ | $=$ | $\u27e8f(x)z-f(z)x,z\u27e9$ | ||

$=$ | $f(x)\u27e8z,z\u27e9-f(z)\u27e8x,z\u27e9$ | |||

$=$ | $f(x){\parallel z\parallel}^{2}-\u27e8x,\overline{f(z)}z\u27e9$ | |||

$=$ | $f(x)-\u27e8x,\overline{f(z)}z\u27e9$ |

which implies

$$f(x)=\u27e8x,\overline{f(z)}z\u27e9.$$ |

The theorem then follows by taking $u=\overline{f(z)}z$.

Uniqueness - Suppose there were ${u}_{1},{u}_{2}\in \mathscr{H}$ such that for every $x\in \mathscr{H}$

$$f(x)=\u27e8x,{u}_{1}\u27e9=\u27e8x,{u}_{2}\u27e9.$$ |

Then $\u27e8x,{u}_{1}-{u}_{2}\u27e9=0$ for every $x\in \mathscr{H}$. Taking $x={u}_{1}-{u}_{2}$ we obtain ${\parallel {u}_{1}-{u}_{2}\parallel}^{2}=0$, which implies ${u}_{1}={u}_{2}$. $\mathrm{\square}$

Title | proof of Riesz representation theorem |
---|---|

Canonical name | ProofOfRieszRepresentationTheorem |

Date of creation | 2013-03-22 17:32:37 |

Last modified on | 2013-03-22 17:32:37 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 5 |

Author | asteroid (17536) |

Entry type | Proof |

Classification | msc 46C99 |