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 \[ \begin{aligned} \int_0^\infty |F(x)|^pdx &= \int_0^\infty \lim_{n \to \infty}|F_n(x)|^pdx \\ &\leq \lim_{n \to \infty} \int_0^\infty |F_n(x)|^pdx \\ &\leq \lim_{n \to \infty}q^p\int_0^\infty |f_n(x)|^pdx \\ &=q^p\int_0^\infty |f(x)|^pdx. \end{aligned} \]