# a subset of $l_2$.

Mathematics Asked by Luiza Silva on September 24, 2020

the exercise is: show that the set $$A={x=(x_n)subset l_2: sum(1+frac{1}{i})x_i^2)leq 1}$$ doesn’t contain an element with norm equal to $$sup{|x|_2,xin A}$$.

My attempt: i showed that $$forall xin A, |x|_2^2=sum x_i^2< sum x_i^2+frac{x_i^2}{i}leq 1$$. And $$sup_{xin A} |x|=1$$. Using that $$l_2$$ is a Hilbert space, is there a way to show that, supposing that exists an element that there is norm equal to $$sup{|x|_2,xin A}$$, find some contradiction? This exercise i found at functional analysis and infinite-dimensional geometry- Marián Fabian.

Consider $$xin A$$. Since for all $$ige 1$$ we have $$x_i^2le (i+1/i)x_i^2$$ we conclude by adding these inequalities that $$Vert xVert_2^2le 1$$, so $$sup_{xin A}Vert xVert_2le 1.$$ Now, if we define $$x^{(k)}inell_2$$ by $$x^{(k)}_k=1/sqrt{1+1/k}$$ and $$x^{(k)}_i=0$$ for $$ine k$$, then $$x^{(k)}in A$$ and $$Vert x^{(k)}Vert_2=frac{1}{sqrt{1+1/k}}le sup_{xin A}Vert xVert_2le 1$$ But since $$k$$ is arbitrary we conclude by letting $$k$$ tend to $$infty$$ that $$sup_{xin A}Vert xVert_2=1.$$

Now, suppose that there exists $$yin A$$ such that $$Vert yVert_2=1$$ then $$sum_{ige1}frac1i y_i^2= sum_{ige1}left(1+frac1i right)y_i^2 -sum_{ige1}y_i^2le1-Vert yVert_2^2=0$$ Thus $$sum_{ige1}frac1i y_i^2=0$$ So $$y_i=0$$ for all $$i$$ and this contradicts the fact that $$Vert yVert_2=1$$. This contradiction proves that there is no $$yin A$$ such that $$Vert yVert_2=1$$. $$qquadsquare$$

Correct answer by Omran Kouba on September 24, 2020

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