Is the Cesaro Operator normal?

Question

The Cesàro operator $T:ℓ_p→ℓ_p$ is defined by $$(Tx)_k=(1/k)\sum_{j=1}^k x_j$$ where $x=(x_j)$. Is this operator normal?

Answer 1

In the case $p = 2$, we first compute $T^*$. Noting that

$$ \langle Tx, y\rangle = \sum_{k} \sum_{j = 1}^k x_j \bar{y}_k / k = \sum_{j} \sum_{k = j}^\infty x_j \bar{y}_k / k $$

we have that

$$ (T^*y)_j = \sum_{k = j}^\infty \frac{y_k}{k} $$

So we have that

$$ (T T^* y)_\ell = \sum_{j = 1}^\ell \sum_{k = j}^\infty \frac{y_k}{k \ell} $$

and

$$ (T^* T x)_\ell = \sum_{k = \ell}^\infty \sum_{j = 1}^k \frac{x_j}{k^2} $$

Testing against the sequence $(1,0,0,\ldots)$ you see the two are clearly not equal. So even in the $\ell_2$ case the operator is not normal.

In the $\ell_p$ case I don't even know what "normal" would mean.