= Solution
The real <Riesz representation theorem> states that for every bounded linear functional $f:H\to\mathbb R$ on a real Hilbert space there is a unique $y\in H$ such that
$$
f(x)=\langle x,y\rangle
\qquad(x\in H),
$$
and $\|f\|=\|y\|$.
If $f=0$, take $y=0$. Otherwise $Y=\ker f$ is a closed proper subspace, so the supplied <orthogonal decomposition by a closed subspace> gives a nonzero $z\in Y^\perp$. Since $f(z)\ne0$, for every $x\in H$,
$$
x-\frac{f(x)}{f(z)}z\in Y.
$$
Taking its inner product with $z$ yields
$$
\langle x,z\rangle
=\frac{f(x)}{f(z)}\|z\|^2,
$$
and therefore
$$
f(x)=\left\langle x,\frac{f(z)}{\|z\|^2}z\right\rangle.
$$
Thus $y=f(z)z/\|z\|^2$ represents $f$. If both $y$ and $y'$ represent it, then $\langle x,y-y'\rangle=0$ for every $x$; taking $x=y-y'$ proves uniqueness. Finally, <Cauchy-Schwarz inequality> gives $\|f\|\leq\|y\|$, while evaluating at $x=y/\|y\|$ when $y\ne0$ gives the reverse inequality. Hence
$$
\boxed{f(x)=\langle x,y\rangle,
\qquad\|f\|=\|y\|.}
$$
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