I am looking at the proof of the following theorem: $$$$ Let $\widetilde{H}$ subspace of $H$,where $H$ is an Euclidean space, and $x \in H$. $y \in \widetilde{H}$ is the optimal approximation of $x$ of $\widetilde{H}$ iff $(x,u)=(y,u) \forall u \in \widetilde{H} \text{ or } (x-y,u)=0\ \forall u \in \widetilde{H}$
I got stuck at the proof of this direction: $\Rightarrow$ $$$$ In my notes they do it like that: $$$$ Let $y \in \widetilde{H}$ is the optimal approximation of $x$. We define $\varphi(λ)=||x-(y+λz)||^{2},z \in \widetilde{H} $ $$\varphi(λ)=||x-y||^{2}-2λ (x-y,z)+λ^2 ||z||^{2}$$ $$$$ We want to find the minimum of $\varphi(λ)$,so we find $\varphi'(λ)=0 \Rightarrow -(x-y,z)+λ||z||^2=0 \Rightarrow λ_{min}=\frac{(x-y,z)}{||z||^2}$ $$min \varphi(λ)=\varphi(λ_{min}) < \varphi(0)=||x-y||^2$$ That can't be true,because the optimal approximation of $x$ of $\widetilde{H}$ is $y$
Could you explain it to me?? Why do we take $\varphi(λ)$ ?What is it?? :/
We want to prove that $y$ is the optimal approximation.
So suppose it is not.
Then that must mean that there is some other vector in $\tilde H$ that is a better approximation. Any vector in $\tilde H$ can be written as $y+\lambda z$ (btw, that should be a $+$ instead of a $-$). So let's suppose that vector is a better approximation.
The function $\varphi(\lambda)$ is the squared distance of $x$ to this other approximation as function of $\lambda$. If we can prove that $\lambda$ must be zero to find the shortest squared distance, we have proven that $y$ is the optimal approximation after all.