# Several ways to prove Hardy's inequality

Suppose \(1 < p < \infty\) and \(f \in L^p((0,\infty))\) (with respect to Lebesgue measure of course) is a nonnegative function, take \[ F(x) = \frac{1}{x}\int_0^x f(t)dt \quad 0 < x <\infty, \] we have Hardy's inequality \(\def\lrVert[#1]{\lVert #1 \rVert}\) \[ \lrVert[F]_p \leq q\lrVert[f]_p \] where \(\frac{1}{p}+\frac{1}{q}=1\) of course.

There are several ways to prove it. I think there are several good reasons to write them down thoroughly since that may be why you find this page. Maybe you are burnt out since it's *left as exercise*. You are assumed to have enough knowledge of Lebesgue measure and integration.

## Minkowski's integral inequality

Let \(S_1,S_2 \subset \mathbb{R}\) be two measurable set, suppose \(F:S_1 \times S_2 \to \mathbb{R}\) is measurable, then \[ \left[\int_{S_2} \left\vert\int_{S_1}F(x,y)dx \right\vert^pdy\right]^{\frac{1}{p}} \leq \int_{S_1} \left[\int_{S_2} |F(x,y)|^p dy\right]^{\frac{1}{p}}dx. \] A proof can be found at here by turning to Example A9. You may need to replace all measures with Lebesgue measure \(m\).

Now let's get into it. For a measurable function in this place we should have \(G(x,t)=\frac{f(t)}{x}\). If we put this function inside this inequality, we see \[ \begin{aligned} \lrVert[F]_p &= \left[\int_0^\infty \left\vert \int_0^x \frac{f(t)}{x}dt \right\vert^p dx\right]^{\frac{1}{p}} \\ &= \left[\int_0^\infty \left\vert \int_0^1 f(ux)du \right\vert^p dx\right]^{\frac{1}{p}} \\ &\leq \int_0^1 \left[\int_0^\infty |f(ux)|^pdx\right]^{\frac{1}{p}}du \\ &= \int_0^1 \left[\int_0^\infty |f(ux)|^pudx\right]^{\frac{1}{p}}u^{-\frac{1}{p}}du \\ &= \lrVert[f]_p \int_0^1 u^{-\frac{1}{p}}du \\ &=q\lrVert[f]_p. \end{aligned} \] Note we have used change-of-variable twice and the inequality once.

## A constructive approach

I have no idea how people came up with this solution. Take \(xF(x)=\int_0^x f(t)t^{u}t^{-u}dt\) where \(0<u<1-\frac{1}{p}\). Hölder's inequality gives us \[ \begin{aligned} xF(x) &= \int_0^x f(t)t^ut^{-u}dt \\ &\leq \left[\int_0^x t^{-uq}dt\right]^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \\ &=\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \end{aligned} \] Hence \[ \begin{aligned} F(x)^p & \leq \frac{1}{x^p}\left\{\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}}\right\}^{p} \\ &= \left(\frac{1}{1-uq}\right)^{\frac{p}{q}}x^{\frac{p}{q}(1-uq)-p}\int_0^x f(t)^pt^{up}dt \\ &= \left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt \end{aligned} \]

Note we have used the fact that \(\frac{1}{p}+\frac{1}{q}=1 \implies p+q=pq\) and \(\frac{p}{q}=p-1\). Fubini's theorem gives us the final answer: \[ \begin{aligned} \int_0^\infty F(x)^pdx &\leq \int_0^\infty\left[\left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt\right]dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dx\int_0^x f(t)^pt^{up}x^{-up-1}dt \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dt\int_t^\infty f(t)^pt^{up}x^{-up-1}dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}\int_0^\infty f(t)^pdt. \end{aligned} \] It remains to find the minimum of \(\varphi(u) = \left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}\). This is an elementary calculus problem. By taking its derivative, we see when \(u=\frac{1}{pq}<1-\frac{1}{p}\) it attains its minimum \(\left(\frac{p}{p-1}\right)^p=q^p\). Hence we get \[ \int_0^\infty F(x)^pdx \leq q^p\int_0^\infty f(t)^pdt, \] which is exactly what we want. Note the constant \(q\) cannot be replaced with a smaller one. We simply proved the case when \(f \geq 0\). For the general case, one simply needs to take absolute value.

## Integration by parts

This approach makes use of properties of \(L^p\) space. Still we assume that \(f \geq 0\) but we also assume \(f \in C_c((0,\infty))\), that is, \(f\) is continuous and has compact support. Hence \(F\) is differentiable in this situation. Integration by parts gives \[ \int_0^\infty F^p(x)dx=xF(x)^p\vert_0^\infty- p\int_0^\infty xdF^p = -p\int_0^\infty xF^{p-1}(x)F'(x)dx. \] Note since \(f\) has compact support, there are some \([a,b]\) such that \(f >0\) only if \(0 < a \leq x \leq b < \infty\) and hence \(xF(x)^p\vert_0^\infty=0\). Next it is natural to take a look at \(F'(x)\). Note we have \[ F'(x) = \frac{f(x)}{x}-\frac{\int_0^x f(t)dt}{x^2}, \] hence \(xF'(x)=f(x)-F(x)\). A substitution gives us \[ \int_0^\infty F^p(x)dx = -p\int_0^\infty F^{p-1}(x)[f(x)-F(x)]dx, \] which is equivalent to say \[ \int_0^\infty F^p(x)dx = \frac{p}{p-1}\int_0^\infty F^{p-1}(x)f(x)dx. \] Hölder's inequality gives us \[ \begin{aligned} \int_0^\infty F^{p-1}(x)f(x)dx &\leq \left[\int_0^\infty F^{(p-1)q}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}} \\ &=\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}}. \end{aligned} \] Together with the identity above we get \[ \int_0^\infty F^p(x)dx = q\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}} \] which is exactly what we want since \(1-\frac{1}{q}=\frac{1}{p}\) and all we need to do is divide \(\left[\int_0^\infty F^pdx\right]^{1/q}\) on both sides. So what's next? Note \(C_c((0,\infty))\) is dense in \(L^p((0,\infty))\). For any \(f \in L^p((0,\infty))\), we can take a sequence of functions \(f_n \in C_c((0,\infty))\) such that \(f_n \to f\) with respect to \(L^p\)-norm. Taking \(F=\frac{1}{x}\int_0^x f(t)dt\) and \(F_n = \frac{1}{x}\int_0^x f_n(t)dt\), we need to show that \(F_n \to F\) pointwise, so that we can use Fatou's lemma. For \(\varepsilon>0\), there exists some \(m\) such that \(\lrVert[f_n-f]_p < \frac{1}{n}\). Thus \[ \begin{aligned} |F_n(x)-F(x)| &= \frac{1}{x}\left\vert \int_0^x f_n(t)dt - \int_0^x f(t)dt \right\vert \\ &\leq \frac{1}{x} \int_0^x |f_n(t)-f(t)|dt \\ &\leq \frac{1}{x} \left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}}\left[\int_0^x 1^qdt\right]^{\frac{1}{q}} \\ &=\frac{1}{x^{1/p}}\left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}} \\ &\leq \frac{1}{x^{1/p}}\lrVert[f_n-f]_p <\frac{\varepsilon}{x^{1/p}}. \end{aligned} \] Hence \(F_n \to F\) pointwise, which also implies that \(|F_n|^p \to |F|^p\) pointwise. For \(|F_n|\) we have \[ \begin{aligned} \int_0^\infty |F_n(x)|^pdx &= \int_0^\infty \left\vert\frac{1}{x}\int_0^x f_n(t)dt\right\vert^p dx \\ &\leq \int_0^\infty \left[\frac{1}{x}\int_0^x |f_n(t)|dt\right]^{p}dx \\ &\leq q\int_0^\infty |f_n(t)|^pdt \end{aligned} \] note the third inequality follows since we have already proved it for \(f \geq 0\). By Fatou's lemma, we have \[ \begin{aligned} \int_0^\infty |F(x)|^pdx &= \int_0^\infty \lim_{n \to \infty}|F_n(x)|^pdx \\ &\leq \lim_{n \to \infty} \int_0^\infty |F_n(x)|^pdx \\ &\leq \lim_{n \to \infty}q^p\int_0^\infty |f_n(x)|^pdx \\ &=q^p\int_0^\infty |f(x)|^pdx. \end{aligned} \]

Several ways to prove Hardy's inequality