I wanted to check my reasoning on proving this statement, and see if anyone had suggestions for other proofs of this fact.
Again, the statement is, if $K$ is a compact operator on a Hilbert space $H$, and if $||Kf|| < ||f|| \ \ \forall f$, then $||K|| < 1$.
Here's my proof. Suppose $||K|| = 1$, i.e. $\sup_{||f|| = 1} ||Kf|| = 1$. Then $$ \sup_{||f|| = 1}<K^*Kf,f> = \sup_{||f|| = 1}<Kf,Kf> = \sup_{||f|| = 1}||Kf||^2 = 1. $$ Then, because $K^*K$ is a compact, self-adjoint operator, by the Rayleigh quotient, $1$ is an eigenvalue of $K$. Thus, there exists $g$ so that $K^*Kg = g$. But then $$||Kg||^2 = <Kg,Kg> = <K^*Kg,g> = <g,g> = ||g||^2$$ and we have a contradiction.
Any problems with this argument?
Also, philosophically, I'd like to prove this in the simplest way possible, in the sense that I'd like to limit the number of intermediate results needed to get between a definition of compact operators (and the Fredholm alternative) and this result. The Rayleigh quotient depends on the existence of a basis of eigenfunctions for $K^*K$, so maybe there's a more direct method?
Thanks in advance for any suggestions/discussion.
Your argument seems flawless to me. I think there is an alternative argument which can generalize the result to reflective Banach spaces. The facts I mainly use are: $(1)$ the closed unit ball in a reflective Banach space is weakly sequentially compact; $(2)$ if $K: X\to Y$ is a compact operator between Banach spaces and $x_n$ weakly converges to $x$ in $X$, then $Kx_n$ strongly converges to $Kx$ in $Y$.
Proof: Let $\{f_n\in H:\|f_n\|=1\}$ be a sequence such that $\lim_{n\to\infty}\|Kf_n\|=\|K\|$. Moreover, we may assume that $f_n$ weakly converges to $f\in H$. Then $\|f\|\le 1$. Since $K$ is compact, $Kf_n$ strongly converges to $Kf$. It follows that $$\|K\|=\lim_{n\to\infty}\|Kf_n\|=\|Kf\|<\|f\|\le 1.\qquad\square$$
Remark: If $H$ is only a Banach space, without the reflective assumption, the conclusion may fail. For example, let $$H=\{f:[0,1]\to\mathbb{R}\mid f \text{ is continuous and } f(0)=0\}$$ endowed with supremum norm and let $K$ be the Volterra operator, i.e. $\forall f\in H$, $$Kf(x)=\int_0^xf(t)dt\,,\quad x\in[0,1].$$ The compactness of $K$ follows from Arzela-Ascoli theorem. Moreover, it is easy to see that $\|K\|=1$(consider $f_n(x)=nx$ on $[0,\frac{1}{n}]$ and $f_n(x)=1$ otherwise) but $\|Kf\|<\|f\|$, $\forall f\ne 0$(using the fact $f(0)=0$).