can a sum of two non-compact operators be compact?

Question

I'm supposed to say, whether an operator $$Tf(t)=f(1-t)$$ can be expressed as $$T=\lambda I-K$$ where $K$ is a compact operator. As $T$ is not compact, I suppose that $K=\lambda I-T$ can't be compact either, but I don't know how to prove it.

Answer 1

So this is how I solved it:

I defined $(f_n)$ as: $$f_n(x)=\chi(2n,2n+1)$$ where $\chi$ is the characteristic function. Follows $$Kf_n(t)=\lambda f_n(t)-f_n(t-1)=\lambda\chi(2n,2n+1)-\chi(2n-1,2n)$$ now lets calculate $||f_n-f_m||_2$: $$||f_n-f_m||_2=\sqrt{\int_{-\infty}^{+\infty}|f_n(x)-f_m(x)|^2}=\sqrt{2\lambda^2+2}$$ which means, that $\forall\lambda\in\mathbb{C}/\{0\}$ we can't choose a convergent subsequence from $(\lambda I-T)(f_n)\ \Rightarrow\ (\lambda I-T)$ can't be compact and $T$ can't be expressed as $\lambda I+K$