Riesz representation theorem (Hilbert space case)

If $\varphi=0$, take $y=0$ and we are done. Otherwise let $N=\ker\varphi=\{x\in H:\varphi(x)=0\}$; since $\varphi$ is continuous and linear, $N$ is a closed proper subspace of $H$. Because $H$ is a Hilbert space, the projection theorem gives an orthogonal decomposition $H=N\oplus N^{\perp}$, and since $N$ is a proper subspace, $N^{\perp}$ contains some $z\ne0$.

For any $x\in H$, consider the vector $u=\varphi(x)z-\varphi(z)x$. Applying $\varphi$ gives $\varphi(u)=\varphi(x)\varphi(z)-\varphi(z)\varphi(x)=0$, so $u\in N$. Since $z\in N^{\perp}$, we get $\langle u,z\rangle=0$, i.e. $\varphi(x)\langle z,z\rangle-\varphi(z)\langle x,z\rangle=0$. Solving for $\varphi(x)$ gives $\varphi(x)=\dfrac{\varphi(z)}{\|z\|^2}\langle x,z\rangle=\left\langle x,\dfrac{\overline{\varphi(z)}}{\|z\|^2}z\right\rangle$, so $y=\dfrac{\overline{\varphi(z)}}{\|z\|^2}z$ represents $\varphi$.

For uniqueness, if $\langle x,y_1\rangle=\langle x,y_2\rangle$ for all $x$, take $x=y_1-y_2$ to get $\|y_1-y_2\|^2=0$, so $y_1=y_2$. For the norm identity, Cauchy–Schwarz gives $|\varphi(x)|=|\langle x,y\rangle|\le\|y\|\|x\|$, so $\|\varphi\|\le\|y\|$; and plugging in $x=y$ gives $\varphi(y)=\|y\|^2$, so $\|\varphi\|\ge|\varphi(y)|/\|y\|=\|y\|$. Together, $\|\varphi\|=\|y\|$.