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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?? :/

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    $\begingroup$ That $y$ is the optimal approximation means that $\varphi$ has its global minimum (not necessarily a unique minimum, consider $z = 0$) in $\lambda = 0$. A minimum is a critical point, hence $\varphi'(0) = 0$, hence ... $\endgroup$ – Daniel Fischer Mar 25 '14 at 14:54
  • $\begingroup$ But..with this proof,have we shown that $(x,u)=(y,u) \forall x \in \widetilde{H} \text{ or that } (x-y,u)=0 \forall x \in \widetilde{H}$ ? $\endgroup$ – evinda Mar 25 '14 at 14:59
  • $\begingroup$ That should be $\forall u\in\widetilde{H}$. Both, $(x,u) = (y,u) \iff (x-y,u) = 0$. $\endgroup$ – Daniel Fischer Mar 25 '14 at 15:09
  • $\begingroup$ But is this proven at the proof that I have written at my first post??? $\endgroup$ – evinda Mar 25 '14 at 15:14
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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.

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    $\begingroup$ What is the definition of an optimal approximation? $\endgroup$ – Klaas van Aarsen Mar 25 '14 at 21:48
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    $\begingroup$ Isn't that what you were asking?? (Confused) $\endgroup$ – Klaas van Aarsen Mar 26 '14 at 21:19
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    $\begingroup$ I'm confused right now.. :/ We suppose that $y \in \widetilde{H}$ is not the optimal approximation of $x \in H$ from $\widetilde{H}$. Any vector in $\widetilde{H}$ can be written as $y+ λz$. We define the function $$\varphi(λ)=||x-(y+λz)||^2=(x-(y+λz), x-(y+λz))=((x-y)-λz, (x-y)-λz)=||x-y||^2-2(x-y, -λz)+||λz||^2=||x-y||^2+2λ(x-y, z)+λ^2||z||^2$$ We want to find the minimum of $\varphi(λ)$, so we find $$\varphi '(λ)=0 \Rightarrow 2(x-y,z)+2λ||z||^2=0 \Rightarrow -(x-y,z)=λ||z||^2 \Rightarrow λ=- \frac{(x-y,z)}{||z||^2}$$ How can I continue?? $\endgroup$ – evinda Mar 31 '14 at 15:46
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    $\begingroup$ Substituting your $\lambda$ in $\varphi(\lambda)$ gives $\varphi(\lambda) = ||x-y||^2 + 3\frac{(x-y, z)^2}{||z||^2}$, which is always at least $||x-y||^2$, which is a contradiction. $\endgroup$ – Klaas van Aarsen Apr 1 '14 at 6:42
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    $\begingroup$ I am looking again at this proof...You told me that: $$\text{Any vector in $\tilde H$ can be written as } y+\lambda z $$ But....why is it like that?? Are $y,z$ elements of the basis of $\widetilde{H}$ ?? $\endgroup$ – evinda Jun 4 '14 at 15:15

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