Here is the proof of the Riesz Lemma.
When the author show that $\| z-y \| \geq \theta$, I don't understand the following step:
$$\dfrac{1}{\| v-y_0 \|} \| v-y_0 - (\| v-y_0 \|)y \| \geq \dfrac{\alpha}{\| v-y_0 \|}$$
How to obtain the above step?
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2$y_0+\Vert v-y_0\Vert y\in Y$ and $\inf_{y\in Y} \Vert v-y\Vert=a$. – David Mitra Oct 27 '15 at 12:08
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Awh I can't believe I mislook this fact.... – Idonknow Oct 27 '15 at 12:10
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$v - y_0 - (||v-y_0||)y$ is of the form $v - y'$ with $y' \in Y$ (as $Y$ is a subspace, and $y, y_0 \in Y$).
The norm of it is thus bounded below by $\alpha$ by the definition of $\alpha$ as $\inf_{y \in Y} ||v - y||$.
Henno Brandsma
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Aren’t they (including summation and scalar product) properties of vector subspace? Is only subspace enough for using them? Thanks in advance – usereb May 10 '18 at 15:17
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1@esrabasar a subspace means vector subspace in the lemma. So $Y$ is closed under addition and scalar multiplication. – Henno Brandsma May 10 '18 at 15:20
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I didn’t know that it means vector subspace. It is same in my lecture notes it says only subspace hence I’m confused while studying. You have clarified very thanks sir – usereb May 10 '18 at 17:12
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@esrabasar the wikipedia page also states it as a vector subspace (follow the link in the first mention of subspace). – Henno Brandsma May 10 '18 at 17:16
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I’ve realized that thanks to your reply. Also I will glance wikipedia. Sincerely – usereb May 10 '18 at 17:19
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