The Hardy inequality for the averaging operator on the half-line

Reduction to bounded compactly supported $f$. For $n\ge 1$ put $f_n:=\min (f,n)\cdot {\mathbf{1}}_{(0,n)}$, a measurable function with $0\le f_n\le n$ supported in $(0,n)$, so that $f_n\uparrow f$ pointwise. Write $F(t):={\int }_0^{t}f$ and $F_n(t):={\int }_0^t{f}_n$; by [F4] applied to the nondecreasing measurable sequence $f_n{\mathbf{1}}_{(0,t)}$ one has $F_n(t)\uparrow F(t)$ for every $t>0$, hence $(F_n(t)\mathrm{/}t{)}^p\uparrow (F(t)\mathrm{/}t{)}^p$ and ${\int }_0^{\mathrm{\infty }}{f}_n^p\uparrow {\int }_0^{\mathrm{\infty }}{f}^p$. Therefore it suffices to prove the inequality for every $f_n$: applying [F4] to both sides then gives ${\int }_0^{\mathrm{\infty }}{t}^{-p}F(t{)}^{p}dt\le (\frac{p}{p-1}{)}^p{\int }_0^{\mathrm{\infty }}{f}^p$, the case $\int f^p=+\mathrm{\infty }$ included.

The bounded compactly supported case: the dual test function and finiteness. Assume now $0\le f\le M$ and $f=0$ outside $(0,T_0)$. Then $F(t)\le \min (Mt,{\int }_0^{\mathrm{\infty }}f)$ for all $t$, so $Hf=F\mathrm{/}t$ is finite on $(0,\mathrm{\infty })$ with $0\le Hf\le M$; put $\phi (t):=(F(t)\mathrm{/}t{)}^{p-1}$, a bounded nonnegative measurable function. Its $L^{p^{\mathrm{\prime }}}$ norm satisfies $\mathrm{\parallel }\phi {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}^{p^{\mathrm{\prime }}}={\int }_0^{\mathrm{\infty }}(F(t)\mathrm{/}t{)}^{p}dt=:I$, and $I<+\mathrm{\infty }$ because $I\le M^p+(1\mathrm{/}(p-1))({\int }_0^{\mathrm{\infty }}f{)}^p$: on $(0,1)$ the bound $Hf\le M$ gives ${\int }_0^1(Hf{)}^p\le M^p$, and on $(1,\mathrm{\infty })$ the bound $Hf\le t^{-1}{\int }_0^{\mathrm{\infty }}f$ gives ${\int }_1^{\mathrm{\infty }}(Hf{)}^p\le (\int f{)}^p{\int }_1^{\mathrm{\infty }}{t}^{-p}dt=(\int f{)}^p\mathrm{/}(p-1)$. Thus $\phi \in L^{p^{\mathrm{\prime }}}(0,\mathrm{\infty })$ and $Hf\in L^p(0,\mathrm{\infty })$.

Sharpness. For $R>1$ set $f_R(t):=t^{-1\mathrm{/}p}{\mathbf{1}}_{(1,R)}(t)$, so $\mathrm{\parallel }{f}_R{\mathrm{\parallel }}_p^p=\log R$. For $1<t<R$ one has $H{f}_R(t)=p^{\mathrm{\prime }}{t}^{-1\mathrm{/}p}(1-t^{-1\mathrm{/}{p}^{\mathrm{\prime }}})$. Given $A>1$ and $R>A$, this gives $\mathrm{\parallel }H{f}_R{\mathrm{\parallel }}_p^p\ge (p^{\mathrm{\prime }}{)}^p(1-A^{-1\mathrm{/}{p}^{\mathrm{\prime }}}{)}^p\log (R\mathrm{/}A)$. Dividing by $\log R$ and letting $R\to \mathrm{\infty }$, then $A\to \mathrm{\infty }$, shows that no constant smaller than $p^{\mathrm{\prime }}$ can bound $\mathrm{\parallel }Hf{\mathrm{\parallel }}_p\mathrm{/}\mathrm{\parallel }f{\mathrm{\parallel }}_p$. Each $H{f}_R$ has finite $L^p$ norm: it vanishes below $1$, and above $R$ it equals $t^{-1}{\int }_1^R{f}_R$.

The duality identity. With $\mathrm{\Phi }(s):={\int }_s^{\mathrm{\infty }}\phi (t)t^{-1}dt$, Tonelli's theorem [F2] applied to the nonnegative measurable function $(s,t)\mapsto f(s)\phi (t)t^{-1}{\mathbf{1}}_{\{s<t\}}$ on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ gives $I={\int }_0^{\mathrm{\infty }}\phi (t)(F(t)\mathrm{/}t)dt={\int }_0^{\mathrm{\infty }}\phi (t)t^{-1}{\int }_0^{t}f(s)dsdt={\int }_0^{\mathrm{\infty }}f(s)\mathrm{\Phi }(s)ds$.

The $L^{p^{\mathrm{\prime }}}$ bound on $\mathrm{\Phi }$. For $s>0$ substitute $t=su$, $u\in (1,\mathrm{\infty })$, to get $\mathrm{\Phi }(s)={\int }_1^{\mathrm{\infty }}\phi (su)u^{-1}du$. Apply Minkowski's integral inequality [F3] to $F(x,u):=\phi (xu)u^{-1}$ on $(0,\mathrm{\infty })\times (1,\mathrm{\infty })$: the hypothesis holds because ${\int }_1^{\mathrm{\infty }}\mathrm{\parallel }\phi (\cdot u)u^{-1}{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}}du=\mathrm{\parallel }\phi {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}{\int }_1^{\mathrm{\infty }}{u}^{-1-1\mathrm{/}{p}^{\mathrm{\prime }}}du=p^{\mathrm{\prime }}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}<+\mathrm{\infty }$. The conclusion gives $\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le p^{\mathrm{\prime }}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$.

The bound for bounded compactly supported $f$. By steps 2.1 and 2.2 and Holder's inequality [F1], $I={\int }_0^{\mathrm{\infty }}f\mathrm{\Phi }\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le p^{\mathrm{\prime }}\mathrm{\parallel }f{\mathrm{\parallel }}_p{I}^{1\mathrm{/}{p}^{\mathrm{\prime }}}$. If $I=0$ there is nothing to prove; otherwise $0<I<+\mathrm{\infty }$ by step 1.2, so dividing by $I^{1\mathrm{/}{p}^{\mathrm{\prime }}}$ gives $I^{1\mathrm{/}p}\le p^{\mathrm{\prime }}\mathrm{\parallel }f{\mathrm{\parallel }}_p$, which is the claimed inequality for $f$.

The interval case. Let $f$ be supported in $(0,T)$ and extend it by zero to $(0,\mathrm{\infty })$; the extension has the same $L^p$ integral and its averaging function equals $t^{-1}{\int }_0^{t}f$ for $t\le T$, so ${\int }_0^T{t}^{-p}({\int }_0^{t}f{)}^{p}dt\le {\int }_0^{\mathrm{\infty }}{t}^{-p}({\int }_0^{t}f{)}^{p}dt\le (p^{\mathrm{\prime }}{)}^p{\int }_0^{\mathrm{\infty }}{f}^p=(p^{\mathrm{\prime }}{)}^p{\int }_0^T{f}^p$, the middle inequality being the general inequality obtained by combining the reduction of step 1.1 with the bounded-case bound of step 3.1.

Conclusion. Step 1.1 reduces the general measurable case to the bounded compactly supported case, which is step 3.1; step 1.3 shows the constant cannot be improved, and step 4.1 discharges the interval form. This proves both displayed inequalities, the sharpness assertion, and the statement for data supported in $(0,T)$.