Given that $T$ is an operator on a Hilbert space $H$ with $\|T\|=1$. The text I am reading says if $(T^*T x,x)<1$ for some unit vector $x$, then we can construct $C$ such that $C$ is small in a neighborhood of $x$ and vanish outside the neighborhood, and if we set $R=I+C$, then $\|T^*TR^2\|=1$, where $I$ is the identity operator on $H$.
I cannot prove this fact myself and need help. Please assume $H$ to be infinite-dimensional (and separable if necessary). Is it direct or trivial? In that case kindly give me some hint. Thanks in advance.
This is maybe more of a hint or answer sketch than an answer, but it's too long to be a comment, so here goes.
From the quote, I infer that this is from Richard Kadison's paper Isometries of Operator Algebras in Annals of Mathematics, 54(2), 325 (doi:10.2307/1969534), specifically part of an argument on page 327. Key context is present in a few sentences that were omitted from the excerpt in the original post (in particular, $x$ is not a unit vector in the Hilbert space on which $T$ acts, and the original source uses the same symbol "$C$" to denote two different things).
As background, at this point in the paper, $T$ is an element of the unit sphere $\mathfrak{S}$ of a $C^*$-algebra $\mathfrak{A}$, where for Kadison a "$C^*$-algebra" is a self-adjoint, norm-closed subalgebra of operators on a Hilbert space (see page 325). Kadison uses $\mathcal{H}$ to denote the Hilbert space that the operators in $\mathfrak{A}$ act on (see the very bottom of page 326). Kadison then states:
In saying "[c]onsider the commmutative, real $C^*$-algebra . . ., and its representing $C(X)$" Kadison is appealing to general theory that if $\mathfrak{A}_0$ denotes the real $C^*$-subalgebra of $\mathfrak{A}$ generated by $I$ and $T^* T$, then there is a compact Hausdorff space $X$ (uniquely determined up to homeomorphism by $T$) and an isomorphism $\Phi: \mathfrak{A}_0 \to C(X)$, where $C(X)$ denotes the algebra of real-valued continuous functions on $X$ with the usual supremum norm. I do not see a citation for this in Kadison's paper, although the details would have been familiar to his audience and the paper of Gelfand and Naimark that he cites elsewhere probably has them (many more modern textbooks would also have them).
When Kadison says "[d]enoting operators and their representing functions by the same symbol" he is saying that he will identify each $A \in \mathfrak{A}_0$ ("operator") with its image $\Phi(A)$ in $C(X)$ ("representing function"), and in particular, that he will not distinguish between them notationally or otherwise refer to them as different things. In particular:
When he calls $T^* T$ "a positive function of norm $1$" he is saying that $\Phi(T^* T)$ is a positive function of norm $1$. In $C(X)$, the $C^*$-algebraic notion of positivity is the usual notion of positivity of functions, and the $C^*$-algebra norm is the supremum norm on the compact space $X$. So here he is saying that $\Phi(T^* T)$ is not just a real-valued continuous function on $X$, but a continuous function on $X$ that takes values in $[0,\infty)$ (i.e., it is a "positive function") and attains the supremum $1$ somewhere on $X$ (i.e., it has "norm 1" in the norm of $C(X)$).
When he refers to $T^* T$ as "tak[ing] a nonzero value less than one" he is talking about a real-number value assumed by the function $\Phi(T^* T)$ somewhere on $X$ (and not, for example, any Hilbert-space-valued "value" assumed by the operator $T^* T$ on any vector in $\mathcal{H}$). He is about to show that $\Phi(T^* T)$, which at this stage in the argument is only known to be continuous function from $X$ to $[0,1]$, is under the given hypotheses actually a continuous function from $X$ to $\{0,1\}$.
When he says "a function $C$ small in a neighborhood of $x$, vanishing outside this neighborhood, and nonzero at $x$ such that if $R = I + C$ and $S = I - C$. . ." he is first talking about an element $c$ of $C(X)$ that is nonzero at $x$, has small supremum norm in $C(X)$, and vanishes outside a neighborhood of $x$, and then looking at the operator $\Phi^{-1}(c) \in \mathfrak{A}_0$ on $\mathcal{H}$ that corresponds to that function under $\Phi$, and considering the operators $I + \Phi^{-1}(c)$ and $I - \Phi^{-1}(c)$.
To preserve the notational distinction that the paper does not, let's write $f \in C(X)$ for the "representing function" of $T^* T$. Because $T^* T$ is a positive operator of norm $1$, $f$ is a positive function of norm $1$ in the usual sense, i.e., $f$ assumes nonnegative real values, and is bounded above by $1$, and attains the value $1$ at some point of $X$. In the language of functions, Kadison then uses the following lemma.
One intuition behind the lemma is look for $c$ in the form of a continuous "bump" that is only nonzero when $f$ is "far away" from $1$, and where the magnitude of "bump" is small enough that $|f(t) (1 \pm c(t))^2|$ stays less than $1$ when $f(t)$ is "far away" from $1$. Indeed, any intuition you would get for this from drawing a picture of an example in the case $X \subseteq [0,1]$ will carry right over to the general case. Presumably Kadison is avoiding exhibiting specific $c$ because there are some essentially arbitrary choices involved in constructing particular examples (all he uses later about $c$ are the properties listed in the conclusion of the lemma).
To get this intuition closer to a construction, fix a number $a$ satisfying $f(x) < a < 1$. By some version of Urysohn's lemma there is a continuous $h: X \to [0,1]$ such that $h(x) = 1$ and $h(t) = 0$ whenever $f(t) \geq a$ (one could be much more concrete about this but I am trying to place this in a general framework where intuition might be clearer). Because $t \mapsto (1 + t)^2$ and $t \mapsto (1 - t)^2$ are both continuous at $0$, given any $k > 1$ there is $\delta$ such that if $|t| \leq \delta$ then $(1 \pm t)^2 < k$. You can then check that e.g. if you take $\delta$ coming from the choice $k = 1/a$, the function $c = \delta h$ has the required properties. [Prove $|f(t) (1 \pm c(t))^2| \leq 1$ in cases, depending on whether $f(t) < a$, where you use the choice of $\delta$, or $f(t) \geq a$, where you use the definition of $h$.]