Apologies if this is a duplicate. And just to note, this is self-study. Let $B$ be a compact self-adjoint operator on a Hilbert space $H$.
We consider the polynomial $p(x) = a_0+ \sum_{i=1}^n a_i x^i$. Consider now the polynomial of $B$, $$ p(B) = a_0 I +\sum_{i=1}^n a_i B^i. $$ I am to show that $p(B)$ be a compact operator for all real values of the coefficients. I believe that this cannot be the case unless $a_0 = 0$. My thinking is that we know that the (positive) powers and sums of compact operators are again compact, so $\sum_{i=1}^n a_i B^i$ is compact. If we assume that $p(B)$ is compact, then $p(B)-\sum_{i=1}^n a_i B^i$ should be compact, but this is $a_0 I$, and the identity operator is not compact (unless H is finite-dimensional), so $p(B)$ cannot be compact... but nevertheless, the lecture notes claim that this is the case, so where am I making a mistake?