# Showing that $0 \in \overline{A(\partial B_1^X)}$ if $X,Y$ Banach with $\dim X = + \infty$ and $A \in \mathcal{L}_c(X,Y)$.

Exercise :

Let $$X,Y$$ be Banach spaces with $$\dim X = + \infty$$ and $$A \in \mathcal{L}_c(X,Y)$$. Show that $$0 \in \overline{A(\partial B_1^X)}$$.

Attempt :

So, since $$A$$ is a Linear Compact operator $$A : X \to Y$$, this means that it "transfers" bounded subsets of $$X$$ to relatively compact sets in $$Y$$ (meaning their closure is compact). The unit ball $$B_1^X$$ in $$X$$ is such a bounded set. That means that $$\overline{A(B_1^X)}$$ is a compact set.

Now, since $$\partial B_1^X = \{ x \in X : \|x\|_X = 1\}$$ does that mean that it is also bounded ?

Then, if that's correct, $$\overline{A(\partial B_1^X)}$$ should be compact.

Since $$A \in \mathcal{L}_c(X,Y)$$ then the image (range) of $$A$$ is separable, but I don't know (or see) how I could get that into the mix here.

How would one actually show that $$0 \in \overline{A(\partial B_1^X)}$$ ?

## 1 Answer

We want to show that there is a sequence of unit vectors $$(x_n)_{n\ge 1}$$ such that $$\lim_{n\to\infty}Ax_n=0$$. Note that this is equivalent to $$0\in \overline{A(\partial B_1^X)}$$. Assume to the contrary that there exists a $$c>0$$ such that for every unit vector $$x$$, $$\|Ax\|\ge c$$, or equivalently, $$\|Ax\|\ge c\|x\|(*)$$ for all $$x\in X$$. Then it follows that $$A$$ is injective and has a closed range. This is because if $$Ax_n\to y$$, then $$\|A(x_n-x_m)\|\to 0$$ as $$n,m\to\infty$$, which in turn implies that $$(x_n)_{n\ge 1}$$ is Cauchy by $$(*)$$. Thus it follows that $$A:X\to A(X)$$ is a bijection between two Banach spaces, so there exists a continuous inverse $$A^{-1}$$ by inverse mapping theorem. Now, observe that $$I_X=A^{-1}A:X\to X$$ should be a compact operator as it is a product of a compact operator and some bounded linear operator. However, since $$X$$ is infinite-dimensional, $$I_X$$ cannot be compact. This leads to a contradiction. So there is $$(x_n)_{n\ge 1}\subset B_1^X$$ such that $$Ax_n \to 0$$ as $$n\to\infty$$.