Computing the best constant in classical Hardy's inequality Classical Hardy's inequality (cfr. Hardy-Littlewood-Polya Inequalities, Theorem 327)
If $p>1$, $f(x) \ge 0$ and $F(x)=\int_0^xf(y)\, dy$ then 
$$\tag{H} \int_0^\infty \left(\frac{F(x)}{x}\right)^p\, dx < C\int_0^\infty (f(x))^p\, dx $$
unless $f \equiv 0$. The best possibile constant is $C=\left(\frac{p}{p-1}\right)^p$. 
I would like to prove the statement in italic regarding the best constant. As already noted by Will Jagy here, the book suggests stress-testing the inequality with 
$$f(x)=\begin{cases} 0 & 0\le x <1 \\ x^{-\alpha} & 1\le x \end{cases}$$ 
with $1/p< \alpha < 1$, then have $\alpha \to 1/p$. If I do so I get for $C$ the lower bound
$$\operatorname{lim sup}_{\alpha \to 1/p}\frac{\alpha p -1}{(1-\alpha)^p}\int_1^\infty (x^{-\alpha}-x^{-1})^p\, dx\le C$$
but now I find myself in trouble in computing that lim sup. Can someone lend me a hand, please?

UPDATE: A first attempt, based on an idea by Davide Giraudo, unfortunately failed. Davide pointed out that the claim would easily follow from 
$$\tag{!!} \left\lvert \int_1^\infty (x^{-\alpha}-x^{-1})^p\, dx - \int_1^\infty x^{-\alpha p }\, dx\right\rvert \to 0\quad \text{as}\ \alpha \to 1/p. $$
But this is false in general: for example if $p=2$ we get 
$$\int_1^\infty (x^{-2\alpha} -x^{-2\alpha} + 2x^{-\alpha-1}-x^{-2})\, dx \to \int_1^\infty(2x^{-3/2}-x^{-2})\, dx \ne 0.$$
 A: It's been a long time since the original post, but here is yet another way to show that $\frac{p}{p-1}$, $1<p<\infty$, is the best constant in Hardy's inequality once it has been established that
$\|Hf\|_p\leq\frac{p}{p-1}\|f\|_p$ where $Hf(x)=\frac1x\int^x_0f(t)\,dt$. This follows a suggestion in Exercise 3.14  in Rudin's book of Real and Complex Analysis.   For $A>1$, define
$$ f_A(x)=x^{-1/p}\mathbf{1}_{(1,A]}(x)$$
A straight forward computation shows that $\|f_A\|^p_p=\log(A)$. On the other hand, the Hardy transform of $f_A$ is
$$ F_A(x):=Hf_A(x)=\frac{p}{p-1}\frac{1}{x}\Big((\min(A,x))^{1-\tfrac{1}{p}}-1\Big)\mathbf{1}_{(1,\infty)}(x) $$
Then
$$
\begin{align*}
\|F_A\|^p_p &= \Big(\frac{p}{p-1}\Big)^p\Big( \int^A_1(x^{-1/p}-x^{-1})^p\,dx + \frac{(A^{\frac{1-p}{p}} - 1)^p}{p-1}\Big)
\end{align*}
$$
We now normalize $f_A$, by the factor $\frac{1}{\|f_A\|_p}$ to obtain
$$\frac{\|F_A\|_p}{\|f_A\|_p} \geq \Big(\frac{p}{p-1}\Big)\left(\frac{\int^A_1(x^{-1/p}-x^{-1})^p\,dx}{\log(A)}\right)^{1/p}\xrightarrow{A\rightarrow\infty}\frac{p}{p-1}$$
From this, it follows immediately that $C:=\frac{p}{p-1}$ is the best constant in Hardy's inequality. In terms of operators, as user user345872 suggested above, the operator norm of the Hardy transform $f\mapsto Hf$ on $L_p((0,\infty),dx)$ is indeed $\frac{p}{p-1}$.
A: This is an extended comment to the answer of Olivier Diaz. Let 
$$
Hf(x):=\frac1 x\int_0^x f(y)\, dy.$$
I computed the limit 
$$
\ell=\lim_{A\to \infty} \frac{ \| Hf_A^\alpha\|_p^p}{\|f_A^\alpha\|_p^p}, $$
where $f_A^\alpha:=x^{-\alpha}\mathbf 1_{\{1\le x\le A\}}$, for $\alpha \in (0, 1)$. 
When $\alpha<1/p$, 
$$
\ell=(1-\alpha)^{-p} +\frac{1-\alpha p}{(1-\alpha)(p-1)}.$$
When $\alpha >1/p$, 
$$
\ell=\frac{\alpha p -1}{(1-\alpha)^p}\int_1^\infty (x^{-\alpha}-x^{-1})^p\, dx.$$
When $\alpha=1/p$, as in Olivier's answer, 
$$
\ell=\left(\frac{p}{p-1}\right)^p.$$
The following picture shows the graph of $\ell$ as a function of $\alpha$ (for $p=4$). The peak occurs at $\alpha=0.25$, that is, $\alpha=1/p$.

The fact that $\ell$ is maximal at $\alpha=1/p$ is to be expected; as Olivier shown in his answer, $f^{1/p}_A$ attains the supremum of the ratio $\|Hg\|_p/\|g\|_p$, in the limit $A\to \infty$. However, this picture also contains another, new, piece of information; it shows that $\alpha=1/p$ is the only value of $\alpha$ that attains such supremum. In other words, among all powers $x^{-\alpha}$, and in a vague sense,

$$
x^{-1/p}\text{ is the only maximizer to the Hardy inequality.}$$

See this great answer of David C. Ullrich.
A: What you need isn't
$$\lim_{\alpha\searrow1/p}\,\left\lvert \int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx - \int_1^\infty x^{-\alpha p }\mathrm dx\right\rvert=0$$
but
$$\lim_{\alpha\searrow1/p}\frac{\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx}{\int_1^\infty x^{-\alpha p }\mathrm dx}=1\;,$$
which is indeed the case, since as $\alpha\searrow1/p$, the integrals are more and more dominated by regions where $x^{-1}\ll x^{-\alpha}$. For arbitrary $b\gt1$ and $1/p\lt\alpha\lt1$, we have
$$
\begin{eqnarray}
\int_1^\infty (x^{-\alpha})^p\mathrm dx
&\gt&
\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx
\\
&\gt&
\int_b^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx
\\
&\gt&
\int_b^\infty (x^{-\alpha}-b^{\alpha-1}x^{-\alpha})^p\mathrm dx
\\
&=&
(1-b^{\alpha-1})^p\int_b^\infty (x^{-\alpha})^p\mathrm dx
\\
&=&
(1-b^{\alpha-1})^pb^{1-\alpha p}\int_1^\infty (x^{-\alpha})^p\mathrm dx
\\
&=&
(b^{1/p-\alpha}-b^{1/p-1})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;.
\end{eqnarray}
$$
Then choosing $b=2^{1/\beta}$ with $\beta=\sqrt{\alpha-1/p}$ yields
$$\int_1^\infty (x^{-\alpha})^p\mathrm dx
\gt
\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx
\gt
(2^{-\beta}-2^{(1/p-1)/\beta})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;.
$$
Since $\beta\to0$ as $\alpha\searrow1/p$, the factor on the right goes to $1$; thus,
$$\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx\sim\int_1^\infty x^{-\alpha p }\mathrm dx\quad\text{as}\quad\alpha\searrow1/p,$$
as required.
A: Incidentally, one can further extend the ideas detailed by Giuseppe Negro and user user345872 above to show that the Hardy transform $T$ as an operator on $L_p(0,\infty)$ ($1<p<\infty$) is not compact. Consider the family of functions $\{\phi_A(x)=A^{1/p}\mathbf{1}_{(0,1/A]}(x): A>0\}$. We have that 
$$\|\phi_A\|_p=1$$
and
$$G_A(x):=(T\phi_A)(x)=\frac1x\int^\infty_0\phi_A = A^{1/p}\Big(\mathbf{1}_{(0,1/A]}(x) + \frac{1}{Ax}\mathbf{1}_{(1/A,\infty)}(x)\Big)$$
Then, for $A<B$ we have
$ |G_A - G_B|\geq \big|A^{1/p -1} -B^{1/p -1}|\frac{1}{x}\mathbf{1}_{(1/B,\infty)}(x)$, 
whence we conclude that
$$\|G_A - G_B\|_p \geq \frac{1}{(1-p)^{1/p}}\left|1-(B/A)^{1-\frac1p}\right|$$
Putting things together, we have that $g_n:=\phi_{2^n}$ defines a sequence in unit ball in $L_p(0,\infty)$ for which  $\{Tg_n:n\in\mathbb{N}\}$ has no convergent subsequence (also in $L_p(0,\infty)$).
A: 
We have the operator $T: L^p(\mathbb{R}^+) \to L^p(\mathbb{R}^+)$ with $p \in (1, \infty)$, defined by$$(Tf)(x) := {1\over x} \int_0^x f(t)\,dt.$$Calculate $\|T\|$.

For the operator $T$ defined above, the operator norm is $p/(p - 1)$. We will also note that this is also a bounded operator for $p = \infty$, but not for $p = 1$.
Assume $1 < p < \infty$, and let $q$ be the dual exponent, $1/p + 1/q = 1$. By the theorem often referred to as "converse Hölder,"$$\|Tf\|_p = \sup_{\|g\|_q = 1}\left|\int_0^\infty (Tf)(x)g(x)\,dx\right|.$$So, assume that$\|g\|_q = 1$,\begin{align*} \left| \int_0^\infty (Tf)(x)g(x)\,dx\right| & \le \int_0^\infty |Tf(x)||g(x)|\,dx \le \int_0^\infty \int_0^x {1\over x}|f(t)||g(x)|\,dt\,dx \\ & = \int_0^\infty \int_0^1 |f(ux)||g(x)|\,du\,dx = \int_0^1 \int_0^\infty |f(ux)||g(x)|\,dx\,du \\ & \le \int_0^1 \left(\int_0^\infty |f(ux)|^pdx\right)^{1\over p} \left(\int_0^\infty |g(x)|^q dx\right)^{1\over q}du \\ & = \int_0^1 u^{-{1\over p}}\|f\|_p \|g\|_q du = {p\over{p - 1}}\|f\|_p.\end{align*}So that gives us that the operator norm is at most $p/(p - 1)$. To show that this is tight, let $f(x) = 1$ on $(0, 1]$ and zero otherwise. We have $\|f\|_p = 1$ for all $p$. We can then compute that$$(Tf)(x) = \begin{cases} 1 & 0 < x \le 1 \\ {1\over x} & x > 1\end{cases}$$and by direct computation, $\|Tf\|_p = p/(p - 1)$ for $p > 1$.
The same example also shows that we can have $f \in L^1$ but $Tf \notin L^1$, so we must restrict to $p > 1$. However, it is straightforward to show that $T$ is bounded from $L^\infty \to L^\infty$ with norm $1$. Note that the range of the operator in that case is contained within the bounded continuous functions on $(0, \infty)$.
