This is a problem i came across studying for a functional analysis course.
Consider a Hilbert space $H$ a non-zero vector $v\in H$ and a bounded linear functional $ f :H \longrightarrow \mathbb{C}$ . The point is to show that $P:H \longrightarrow H $ with $$Pu=f(u)v$$ is an orthogonal projection iff $f(v)=1$.
Ok the one direction is easy.Assuming that $P$ is a projection we only need idempotence to prove that $f(v)=1$.For the other direction i thought that i could coclude that the norm of $P$ equals to $1$. Considering a $y=\frac{v}{||v||}\in S^{1}_{H}$ (the surface of the unit sphere in H) we have that $||Py||=\frac{1}{||v||}|f(v)|||v||=1$ . So $$||P||\geq 1$$Now i thought that assuming $||P||>1$ would somehow contradict that $f$ is bounded but i can't get there.
Maybe proving another equivalent condition of $P$ being a projection like orthogonality of $kerP,imP$ or $P=P^*$ would be easier.
Any thoughts would be great
Suppose $v,w$ are orthogonal unit vectors, and let $f(u)=(u,v+w)$. Then $f$ is a continuous linear functional and $f(v)=1$. So $Pv=f(v)v=v$. In order for $P$ to be an orthogonal projection, it is necessary and sufficient that $(z-Pz)\perp \mathcal{R}(P)$, where $\mathcal{R}(P)$ is the range of $P$. Because the range of $P$ is $[\{ v\}]$, which is the subspace generated by the vector $v$, that means $P$ is an orthogonal projection iff $(z-Pz)\perp v$ for all $z\in H$. However, \begin{align} (v+w-P(v+w),v) &=((v+w)-(v+w,v+w)v,v)\\ &=1-(v+w,v+w) \\ &= 1-2\ne 0. \end{align} So you are missing some statement in your problem, or the problem appears to me to be wrong.