On $\mathcal{C}([0,1])$, whether the operator $(\Lambda f)(x)=xf(x)$ is compact.

Question

On the Banach space $\mathcal{C}([0,1])$, whether the operator $(\Lambda f)(x)=xf(x)$ is compact.

We use the following definition of compact operator.

A bounded linear operator $\Lambda:X \to Y$ is compact if, for every bounded sequence $(x_n)_{n\ge1}$ of points in $X$, there exists a subsequence $(x_{n_j})_{j \ge 1}$ such that $\Lambda x_{n_j}$ converges.

Intuitively, I think it is not compact, but I cannot find a sequence as a counterexample.

Can someone give a counterexample or some hints of the proof?

Answer 1

Try letting each $x_n$ be a bump function of norm $1$ supported in $[1-\frac{1}{n},1-\frac{1}{n+1}]$.

Answer 2

Consider $f_n(x)=x^n$, it is bounded on $[0,1]$, $lim_nxf_n(x)=x^{n+1}=0$ if $x<1$ and $lim_nxf_n(1)=1$ so $lim_nxf_n$ cannot have a converging subsequence.

Answer 3

A different ind of argument:

The spectrum of a compact operator consists only of eigenvalues except for $0$, but $\Lambda$ has spectrum $[0,1]$ (because $(\Lambda-\lambda) f(\lambda)=0$ for all $f\in C([0,1])$, $\lambda\in [0,1]$) and none of the spectral values is an eigenvalue (because $(\Lambda-\lambda)f=0$ implies $f(x)=0$ for $x\neq \lambda$, hence $f=0$ by continuity).