*EDIT (28.11.2013).*
This math.SE [question about Cesaro operators](https://math.stackexchange.com/questions/119964/cesaro-operator-is-bounded-for-1p-infty) should also be of interest (it discusses upper bounds; Noam has produced lower bounds below)




---


Federico's recent question [Norm of upper triangular matrix of all ones](https://mathoverflow.net/questions/72361/2-norm-of-the-upper-triangular-all-ones-matrix)
motivated me to ask this question about the norm of a matrix that is, in a sense, dual to the all ones upper triangular matrix.


Let $C\_n$ be the $n \times n$ matrix (transpose of the Cesàro matrix)


$$
[C\_n]\_{ij} = \begin{cases}
 1/j & i \le j\\\\
 0 & i > j
 \end{cases}
$$


What is $\|C\_n\|\_2$? 


Notes: 


1. From results on Cesàro operators, it seems to follow that $\|C\_n\| \le 2$ for all $n \ge 1$.
2. It is also worth noting that $\|C\_n\|\_1=1$ and $\|C\_n\|\_{\infty} = H\_n$, where $H\_n$ denotes the $n$-th Harmonic number.
3. Computing $\frac{e^TC\_n^TC\_ne}{e^Te}=2-H\_n/n$ (e is the all ones vector) we obtain a simple lower-bound on $\|C\_n\|\_2$.



Norm of Cesaro matrix
Here there is likely no closed form for $\|C\_n\|\_2$, but I can show $\|C\_n\|\_2 > 2 - O(1/\log n)$; given that also $\|C\_n\|\_2 \leq 2$ (is this easy to prove?) it follows that $\|C\_n\|\_2 \rightarrow 2$ as $n \rightarrow \infty$.


In general we can consider, for fixed $p \in [1,\infty]$, the $l^p$-operator norm of the Cesàro transformation $C\_n$ that takes any finite sequence $(a\_1,a\_2,\ldots,a\_n)$ to its sequence of averages $j^{-1} \sum\_{i=1}^j a\_i$. [If I've kept my $i$'s and $j$'s straight, that's the action of $C\_n$ on *row vectors*, so this operator norm is what you'd call $\|C\_n\|\_q$ where $p^{-1} + q^{-1} = 1$; fortunately for $p=2$ the $l^p$ norm is self dual.] I claim that this operator norm is at least $p - O(1/\log n)$. Indeed Let $v\_n(p)$ be the sequence whose $j$-th term is $j^{-1/p}$, so $\|v\_n\|\_p^p = H\_n$. Then the image of $v\_n(p)$ under $C\_n$ differs from $p v\_n(p)$ by a vector of $l\_p$-norm $O(1)$, because its $j$-th entry is between $$j^{-1} \int\_1^j \phantom. x^{-1/p} \phantom. dx = p \cdot (j^{-1/p} - j^{-1})$$ and $$j^{-1} \int\_0^j \phantom. x^{-1/p} \phantom. dx = p \phantom. j^{-1/p}.$$ We have thus found a nonzero vector whose norm is multiplied by $p - O(1/\log n)$, which proves the claimed lower bound.


I don't expect that $\|C\_n\|\_2$ has a closed form, because it is a root of the characteristic polynomial of $C\_n^{\rm T} C\_n^{\phantom.}$, and these characteristic polynomials have maximal Galois group for small $n$: running the **gp** code



```
C(n) = matrix(n,n,i,j,(i<=j)/j)
S2(M) = charpoly(M*M~)
P(n)=S2(C(n))
for(n=1,11,print(polgalois(P(n))))

```

I found that the Galois group is $S\_n$ for each $n \leq 11$. The norm $\|C\_n\|\_2$ can then also be computed (as long as $n$ is small enough that the computation terminates in reasonable time) by telling **gp**



```
N(n) = sqrt(vecmax(real(polroots(P(n)))))

```

and the convergence to $2$ seems slow enough that $O(1/\log n)$ might well be the correct error estimate.


Curiously it seems that the action of $C\_n$ on the *column vector* $(a\_n(2))^{\rm T}$ comes rather close to the actual $l^2$-operator norm. For example, when $n=40$ this yields the lower bound $1.5520\ldots$ on the norm (which is actually $1.5594\ldots$), while the row vector gets only $1.4201\ldots$. If this persists as $n \rightarrow \infty$ it should yield a much better estimate on $\|C\_n\|\_2$.


