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 Jun 9 comment Norm of conjugate Hardy operator Oh yes. But I'm interested in the norm of $T^*$ as a linear operator $l^p \rightarrow l^p$. Is the following correct? I'm interested in the norm of $T^*: l^p\rightarrow l^p$. But this is the same as the norm of $(T^*)^*=T: (l^p)^*\rightarrow (l^p)^*$. But $(l^p)^*\tilde{=} l^{p'}$. This norm is known to be $\frac{p'}{p'-1}=p$. So the answer $\lVert T^* \lVert=p$ is correct. Jun 9 comment Norm of conjugate Hardy operator @LeonidKovalev Sorry didn't want to post this as an answer. Your argument obviously doesn't work for $T$ on $l^{\infty}$: $\lVert T\lVert=1$ whereas $\lVert T^* \lVert =\infty$ since for $y=(1,1,\dots)$ we have $(T^* y)_1=\sum_{k=1}^{\infty} \frac{1}{k}$. What am I missing? Jun 9 awarded Student Jun 9 awarded Editor Jun 9 comment Norm of conjugate Hardy operator Probably I don't understand how the argument applies to spaces without scalar product (as $l^p$ for $p\not = 2$). Can you explain? Jun 9 comment Norm of conjugate Hardy operator I don't think this is true. I did some computation in Mathematica using the truncated haromonic series $x_k=\frac{1}{k^{1/p}}$: Table[{p, N[Sum[Sum[1/k^(1 + 1/p), {k, n, 500}]^p, {n, 1, 500}]^(1/p)/ Sum[1/k, {k, 1, 500}]^(1/p)]}, {p, 1, 5}]. The result was: {{1, 1.}, {2, 1.70325}, {3, 2.27878}, {4, 2.75247}, {5, 3.14595}} which makes something like $\lVert T'\lVert=p$ more plausible. Jun 9 comment Norm of conjugate Hardy operator What do you mean by "norm of the form"? This doesn't mean that $\lVert T' \lVert=\lVert T \lVert$ I guess. My intuition tells me that $\lim_{p\rightarrow \infty} \lVert T' \lVert=\infty$ whereas for $T$ the limit is clearly 1. Jun 9 comment Norm of conjugate Hardy operator Thank you! I thought this argument just applied for $p=2$. Does this help in any way with the computation of the norm? Jun 9 revised Norm of conjugate Hardy operator edited title Jun 9 asked Norm of conjugate Hardy operator