Operator norm of translated operator

Question

Suppose $P$ is a compact operator in $L^2([0,1])$. Assume it's contractive in the sense that $\|P\|_{\mathrm{op}}<1$. For some $f$ in $L^2([0,1])$, define the operator

\begin{equation}Tg(x):= f(x) +Pg(x).\end{equation}

Is it true that $T$ remains a contraction? From my initial work, applying the triangle inequality to the right hand side of the definition suggests the norm of $f$ plays some role.

\begin{equation}\|Tg(x)\|_{\mathrm{op}} = \sup_{g \in L^2[0,1]} \frac{\|f(x) +Pg(x)\|_{L^2}}{\|g\|_{L^2}}.\end{equation}

However, intuitively it seems like it should be true as $g$ isn't being "stretched" any more than $P$.

Answer 1

If by contractive you mean in the metric space sense (i.e. $f$ is contractive iff there is a $\lambda<1$ such that for all $x,y$ we have $d(f(x),f(y)) \leq d(x,y)$), then your operator $T$ is indeed still contractive: For any functions $g,h\in L^2([0,1])$ we have $$d(Tg,Th) = ||(f+Pg)-(f+Ph)|| = ||Pg-Ph|| = d(Pg,Ph) \leq \lambda d(g,h),$$ where $\lambda$ is a contraction factor for $P$.

Answer 2

Let's make a simple example. Let $P$ be the halving operator, $g$ be the constant $1$ function, and $f$ be the constant $2$ function. Then $$ T g = f + P g = 2 + \frac{1}{2} 1 = \frac{5}{2} > ||g|| = 1 \text{.} $$

The intuition is that if $f$ is "big", $f$ controls the norm of $Tg$. So then if $g$ is "small" compared to $f$, $T$ can't be contracting (in the sense expressed by the supremum expression you have written).