Prove that $\Lambda$ is not compact

Question

On the Banach space $X=C[0,1]$, consider the linear operator $\Lambda:X\rightarrow X$ defined by $$ (\Lambda f)(0)=f(0),(\Lambda f)(t)=\frac{1}{t}\int_0^tf(s)ds,\forall t>0 $$ Prove that $\Lambda$ is not compact.

My idea: Assume that $\Lambda$ is compact, I am trying to find counterexample. First, let $x_n(t)=t^n$, the norm of $x_n$ is 1, I hope that $\{\Lambda x_n\}$ is not pre-compact. Unfortunely, $\lim_{n\to \infty}$ $|\Lambda x_n|=0$. Then I do not know what to do.

Answer 1

Overkill: Let $f_\alpha (t) = t^\alpha$ for $\alpha \ge 0$, then

$$ \Lambda f_\alpha (t) = \frac{1}{\alpha+1}t^\alpha \Rightarrow \Lambda f_\alpha = \frac{1}{\alpha +1} f_\alpha$$

Thus the point spectrum of $\Lambda$ contains $(0,1]$, this is impossible as compact operators has discrete spectrum.

Answer 2

Take $x_n(t)=\sin(nt)$. Then $\{\Lambda x_n,n\geqslant 1\}$ is not equi-continuous as $\Lambda_n\left(\frac 1n\right)-\Lambda_n\left(0\right)=1-\cos(1)$.