Is infinity norm submultiplicative? What about its power?

Question

Here is the setting

$X$: Positive semi-definite;

Non-negative entries;

$||X||_\infty = \lambda$ (largest absolute row sum is $\lambda$).

$G$: Diagonal matrix;

Non-negative diagonal entries $g_i$;

All $g_i \leq \tau$ (for some positive $\tau$).

$XG$:

all $XG$’s eigenvalue are within unit circle.

I want to bound the largest element of $(XG)^2$.

Two approach:

1. Using submultiplicative: $||XG||_\infty \leq ||X||_\infty \cdot ||G||_\infty \leq \lambda\tau$

so $||(XG)^2||_\infty = ||XG \cdot XG||_\infty \leq ||XG||_\infty \cdot ||XG||_\infty \leq (\lambda\tau) \cdot (\lambda\tau) = \lambda^2 \tau^2$

2. Using matrix definition: $|c_{ij}| \leq \sum_{k=1}^n |x_{ik}| \cdot |x_{kj}| \cdot |g_k| \cdot |g_j| \leq \sum_{k=1}^n \lambda \cdot \lambda \cdot \tau \cdot \tau \leq n \lambda^2 \tau^2$

Obviously, something must be wrong/sloppy. Is $\lambda^2 \tau^2$ the stricter bound? If indeed correct and stricter, can we also derive that using matrix definition? Appreciated!

Answer 1

What about this? Using matrix definition. To find $|(XG)^2|_{max}$, the largest element of $(XG)^2$, we can express $(XG)^2$ as:

$(XG)^2$ = XG * XG = X * G * X * G

Let's consider the $(i, j)$ th element of $(XG)^2$, denoted as $((XG)^2)_{ij}$. $((XG)^2)_{ij}$ = $Σ_k Σ_l (X_{ik} * G_{kk} * X_{lj} * G_{ll})$

Since $G$ is a diagonal matrix, $G_{kk} = g_k$ and $G_{ll} = g_l$, where $g_k$ and $g_l$ are the diagonal entries of $G$. Therefore, $((XG)^2)_{ij}$ = $Σ_k Σ_l (X_{ik} * g_k * X_{lj} * g_l)$

Taking the absolute value and using the triangle inequality, we get: $|((XG)^2)_{ij}|$ <= $Σ_k Σ_l |X_{ik}| * |g_k| * |X_{lj}| * |g_l|$

Since X has non-negative entries, $|X_{ik}| = X_{ik} and |X_{lj}| = X_{lj}$. Also, |g_k| <= t and |g_l| <= t (given). Therefore, $|((XG)^2)_{ij}|$ <= $Σ_k Σ_l X_{ik} * t * X_{lj} * t$ = $t^2$ * $(Σ_k X_{ik})$ * $(Σ_l X_{lj})$

Using the property that the infinity norm of X is λ, we have: $Σ_k X_{ik}$ <= λ and $Σ_l X_{lj} <= λ$ Substituting this in the above inequality, we get: $|((XG)^2)_{ij}|$ <= $t^2$ * λ * λ = $(λ * t)^2$

Any flaw in this proof?