I want to know if we can have some bound of $\| I - \boldsymbol{1}\boldsymbol{p}^T \|_2$, where $I$ denotes the identity matrix of size $k$, and $\boldsymbol{p} \in \mathbb{R}^K$ is some known vector with non-negative entries.
An obvious results can be obtained by using triangular inequality, which yields $$\| I - \boldsymbol{1}\boldsymbol{p}^T \|_2 \leq 1 + K \| \boldsymbol{p} \|,$$ but such bound is not quite tight when $\boldsymbol{p} = \frac{1}{K} \boldsymbol{1}$, under which left-hand-side gives 1 but right-hand-side gives $1+1=2$.
So I wonder if we can leverage some special structure of induced 2-norm to obtain a tighter bound. Thanks!
You can compute the sup norm of your operator exactly.
Let
$$A = I - |1\rangle\langle p|$$
The sup norm of $A$ is the square root of the maximal eigenvalue of $A^\dagger A$:
$$ A^\dagger A = I -|1\rangle\langle p| - |p\rangle\langle 1| + K |p\rangle\langle p|$$
Now the above matrix is one outside the span of $|1\rangle, |p\rangle$. So you can compute the matrix of the above on the vectors $|1\rangle, |p\rangle$.
You get (if I didn't make any mistake)
$$[A^\dagger A] = \left(\begin{array}{cc} 1-\left\Vert p\right\Vert _{1} & -\left\Vert p\right\Vert _{2}^{2}\\ K\left(\left\Vert p\right\Vert _{1}-1\right) & 1-\left\Vert p\right\Vert _{1}+K\left\Vert p\right\Vert _{2}^{2} \end{array}\right) \, , $$
where $\left\Vert p\right\Vert_1 = \sum_j p_j $. The eigenvalues of the above matrix are
$$ \lambda_\pm = \frac{1}{2} \left ( 2(1-\Vert p\Vert_1) +K \Vert p\Vert_2^2 \pm \Vert p\Vert_2 \sqrt{K (4-4 \Vert p\Vert_1 +K \Vert p\Vert_2^2)} \right) $$
Remember also that $A^\dagger A$ is 1 outside $\mathrm{span}(|1\rangle, |p\rangle)$. So your supremum norm is
$$ \Vert A \Vert = \max \left \{1,\sqrt{\lambda_+} \right \}$$