I need some help in the follow question

The question is about the quotient norm:

"True or false: The infimum in the definition of the quotient norm: || [x]|| = inf m∈M ||x − m|| is always attained."

I saw this example in the internet but I don’t understand something:

the example was let $X=l^1$ and let $A=\{x \in l^1 | \sum_{n=1}^{\infty} {\frac{n}{n+1} x_n}=0 \}$ and we try to look at the unit vector $e_1$.

in the internet it clear that the $|| [e_1] ||_1 =0.5$ but I dont know why it is correct

thank to all

I get my example from this blog : http://calculus7.org/2012/03/11/nearest-point-projection-part-i-eschewing-transliterations/


The author wrote

One can check that

the distance is $\frac12$. That doesn't (necessarily) mean it's obvious.

Let's see what we can find. For $a = 0$, we obviously have $\lVert e_1 - a\rVert_1 = 1$, so let's try to find something closer. Write

$$a = c_1\cdot e_1 - \sum_{k=2}^\infty c_k\cdot e_k,$$

where not all $c_k = 0$.

Then to have $a\in A$, we must have

$$\sum_{k=2}^\infty \frac{k}{k+1} c_k = \frac12 c_1,$$

and we have

$$\begin{align} \lVert e_1 - a\rVert_1 &= \lvert 1-c_1\rvert + \sum_{k=2}^\infty \lvert c_k\rvert\\ &> \lvert 1-c_1\rvert + \sum_{k=2}^\infty \frac{k}{k+1}\lvert c_k\rvert\tag{1}\\ &\geqslant \lvert 1-c_1\rvert + \left\lvert \sum_{k=2}^\infty \frac{k}{k+1}c_k\right\rvert\\ &= \lvert 1-c_1\rvert + \frac12\lvert c_1\rvert. \end{align}$$

Now it is easy to see that the last lower bound is minimised for $c_1 = 1$, with

$$\lVert e_1 -a\rVert_1 > \frac12.$$

So we need to see that we can come arbitrarily close to a distance of $\frac12$.

Well, for $a_k = e_1 - \dfrac{k+1}{2k}\cdot e_k,\; k \geqslant 2$, we have $a_k \in A$ and

$$\lVert e_1 - a_k\rVert_1 = \left\lVert \frac{k+1}{2k}e_k\right\rVert = \frac{k+1}{2k} = \frac12 + \frac{1}{2k} \to \frac12,$$

so indeed

$$\inf_{a\in A} \lVert e_1 - a\rVert_1 = \frac12,$$

and by $(1)$, the distance is not attained.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.