Hardy's inequality

Remarks

Proof idea for the integral form: substitute $t = s x$ to get $F(x) = \int_0^1 f(s x) \, ds$, an average of the dilates of $f$. By Minkowski's integral inequality, the continuous form of the Minkowski inequality, $\norm{F}_p \le \int_0^1 \norm{f(s \, \cdot)}_p \, ds$. Since $\norm{f(s \, \cdot)}_p = s^{-1/p} \norm{f}_p$ and $\int_0^1 s^{-1/p} \, ds = \frac{p}{p-1}$, the result follows.

The constant $\big(\frac{p}{p-1}\big)^p$ is the best possible in both forms.

The hypothesis $p > 1$ is needed. For $p = 1$ and $f$ the indicator function of $(0, 1]$, one gets $F(x) = 1/x$ for $x > 1$, so the left side is infinite while the right side is $1$.

Going deeper

For a student

In operator language, the averaging operator $f \mapsto F$ is bounded on the Lp space $L^p(0, \infty)$ for $1 < p < \infty$, with norm $\frac{p}{p-1}$. It is a basic example of a bounded integral operator, and it is used in Sobolev-type estimates and in the theory of weighted inequalities. Letting $p \to \infty$ after replacing $a_n$ by $a_n^{1/p}$ leads to Carleman's inequality.