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Homeproof of Riesz representation theorem

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# proof of Riesz representation theorem

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

Suppose now $f\neq 0$, i.e. $Kerf\neq\mathcal{H}$.

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

$\mathcal{H}=Kerf\oplus(Kerf)^{{\perp}}$ |

and as $Kerf\neq\mathcal{H}$ we can find $z\in(Kerf)^{{\perp}}$ such that $\|z\|=1$.

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

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

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

$\displaystyle\quad\quad\quad\quad\quad\quad 0$ | $\displaystyle=$ | $\displaystyle\langle f(x)z-f(z)x,z\rangle$ | ||

$\displaystyle=$ | $\displaystyle f(x)\langle z,z\rangle-f(z)\langle x,z\rangle$ | |||

$\displaystyle=$ | $\displaystyle f(x)\|z\|^{2}-\langle x,\overline{f(z)}z\rangle$ | |||

$\displaystyle=$ | $\displaystyle f(x)-\langle x,\overline{f(z)}z\rangle$ |

which implies

$f(x)=\langle x,\overline{f(z)}z\rangle\;.$ |

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

Uniqueness - Suppose there were $u_{1},u_{2}\in\mathcal{H}$ such that for every $x\in\mathcal{H}$

$f(x)=\langle x,u_{1}\rangle=\langle x,u_{2}\rangle.$ |

Then $\langle x,u_{1}-u_{2}\rangle=0$ for every $x\in\mathcal{H}$. Taking $x=u_{1}-u_{2}$ we obtain $\|u_{1}-u_{2}\|^{2}=0$, which implies $u_{1}=u_{2}$. $\square$

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46C99*no label found*

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