Bounded linear operator commuting with every compact operators

Question

Let $A$ be a bounded linear operator on the Banach space $X$. Assuming that $AK = KA$ for every compact operator $K$, how do I show that $A$ must be a scalar multiple of the identity, i.e., we have $A = \lambda I$ for some number $\lambda$.

So far I attempt to solve this using Schur's lemma but I can't reason that $A$ will have an eigenvalue in the first place for the lemma to applies (the book where I got the problem from doesn't even specify if the field is real or complex).

Any hint or help is highly appreciated.

Answer 1

I think we can just play around with some special compact operators, unless I'm missing some subtlety. If $x$ and $y$ are linearly independent, then their span is complemented in $X$, so that $X \cong span\{x, y\} \oplus V$ for some closed subspace $V$ of $X$. The projection $P$ along $V$ onto $span\{x, y\}$ is then bounded. Define a map $B: P(X) \rightarrow X$ that swaps $x$ with $y$ and extend by linearity. Then $K_{x,y} = BP$ is continuous and of finite rank (since its image lives in $P(X)$). So it's compact. [I've suppressed in the notation the fact that $K_{x,y}$ depends on a choice of complementary subspace $V$.]

Claim: If there's an $x$ such that $Ax = \lambda x$, then $A=\lambda I$.

Proof: If $x$ and $y$ are linearly independent with $x$ as above, then we have $$Ay=A(K_{x,y}x)=K_{x,y}(Ax)=K_{x,y}(\lambda x)=\lambda y.$$

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Now suppose, by contradiction, that we can find an $x$ such that $x$ and $Ax$ are linearly independent (i.e. $x$ is not an eigenvector for $A$). Then we must have $$x = K_{x, Ax}(Ax) = A(K_{x, Ax}x) = A(Ax)=A^2x,$$ so $x+Ax$ is an eigenvector for $A$. But by our claim, if $A$ has an eigenvector, then $A=\lambda I$, contradicting linear independence of $x$ and $Ax$. So all vectors are eigenvectors, and, by our claim, $A=\lambda I$ for some $\lambda$.

Answer 2

Suppose that $Ax$ is not a scalar multiple of $x$ for some $x \in X$. Then $x\ne 0$, and there is a linear functional $\phi$ defined on the linear span of $\{x,Ax\}$ such that $\phi(x)=1$ and $\phi(Ax)=0$. Such a linear functional is continuous on the two-dimensional subspace spanned by $Ax,x$ and, so, by the Hahn-Banach Theorem, extends to a continuous linear functional $\tilde{\phi}$ on all of $X$. Then $Ky = \tilde{\phi}(y)x$ is a compact linear operator. By assumption $$ KAx = \tilde{\phi}(Ax)x= 0 $$ must equal $$ AKx = \tilde{\phi}(x)Ax = Ax. $$ This is a contradiction because $Ax = 0$ is a scalar multiple of $x$. So, for all $x \in X$, there exists a scalar $\lambda_{x}$ such that $Ax=\lambda_{x}x$.

If $x_{1},x_{2}$ are linearly-independent vectors, there there are bounded linear functionals $\phi_{1},\phi_{2}$ on $X$ such that $\phi_{j}(x_{k})=\delta_{k,j}$. Define $Kx=\phi_{1}(x)x_{2}+\phi_{2}(x)x_{1}$. Then $Kx_{1}=x_{2}$ and $Kx_{2}=x_{1}$. So, $$ AKx_{1} = Ax_{2}=\lambda_{x_{2}}x_{2},\\ KAx_{1} = K\lambda_{x_{1}}x_{1} = \lambda_{x_{1}}x_{2}, $$ which is enough to force $\lambda_{x}=\lambda$ for all $x$. So $A=\lambda I$.