For $0<p<1$ the $H^p$ functional is a quasi-norm, and $H^p$ is a quasi-Banach space

Assume Countable Choice. Fix $n\ge 1$, $0<p<1$ and an admissible kernel $\phi \in \mathcal{S}({\mathbb{R}}^n)$ with $\int \phi \mathrm{\ne }0$ as in The real Hardy space $H^p$ defined by a radial maximal function. Its functional is $p$-subadditive, $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{H^p}^p\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p+\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}^p,$ and fails the ordinary triangle inequality for a pair of elements of this same $H^p$. It also satisfies $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{H^p}\le 2^{1\mathrm{/}p-1}(\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}+\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p})\mathrm{.}$ Consequently $d(f,g)=\mathrm{\parallel }f-g{\mathrm{\parallel }}_{H^p}^p$ is a translation-invariant metric on $H^p$ under which $H^p$ is complete. Thus $H^p$ is a quasi-Banach space. No statement is made identifying $H^p$ with the dual of, or a dual of, a Banach space when $p<1$, and no Banach-space duality theorem is applied to $H^p$ below $p=1$ on this page; the only duality statement here is $H^p=L^p$ for $p>1$.

The maximal operator is pointwise sublinear: $M_{\phi }^0(f+g)\le M_{\phi }^{0}f+M_{\phi }^{0}g$. Since $(u+v{)}^p\le u^p+v^p$ for $u,v\ge 0$ and $0<p<1$, integration gives the stated $p$-subadditivity. The displayed quasi-triangle inequality follows as well because concavity of $r\mapsto r^p$ gives $a^p+b^p\le 2^{1-p}(a+b{)}^p$ for $a,b\ge 0$; take $p$-th roots after $\mathrm{\parallel }f+g{\mathrm{\parallel }}_{H^p}^p\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p+\mathrm{\parallel }g{\mathrm{\parallel }}_{H^p}^p$.

Here is an $H^p$-specific witness that the ordinary triangle inequality fails; the proof does not use the later uniform atom estimate. Put $s=\lfloor n(1\mathrm{/}p-1)\rfloor$ and $\theta (u)=\{\begin{array}{ll}{e}^{-1\mathrm{/}(1-u^2)},& \mathrm{\mid }u\mathrm{\mid }<1,\\ 0,& \mathrm{\mid }u\mathrm{\mid }\ge 1,\end{array}\rho (x)={\prod }_{j=1}^n\theta (x_j),a={\mathrm{\partial }}_1^{s+1}\rho \mathrm{.}$ Here $\theta (u)=\beta (1-u^2)$ for the standard flat function of The standard flat function; The standard flat function is smooth and flat at zero establishes smoothness through the endpoints. Thus $a$ is a smooth function on ${\mathbb{R}}^n$ supported in $[-1,1{]}^n$. It is nonzero: otherwise each one-variable section of $\rho$ would have $(s+1)$st derivative zero, hence would be a polynomial of degree at most $s$ by repeated Newton-Leibniz, impossible for its nonzero compact support. Repeated one-variable integration by parts (obtained from the product rule and Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative) has no boundary terms and gives $\int x^{\alpha }a(x)dx=0$ for every multi-index $\mathrm{\mid }\alpha \mathrm{\mid }\le s$: indeed ${\alpha }_1\le s$, so ${\mathrm{\partial }}_1^{s+1}{x}^{\alpha }=0$. Let $m=M_{\phi }^{0}a$. For every $x$, the convolution estimate gives $m(x)\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }\phi {\mathrm{\parallel }}_1$. For $\mathrm{\mid }x\mathrm{\mid }>2\sqrt{n}$, Taylor's formula in the variable $y$, with the cancellation moments through order $s$ removed, gives for every $t>0$ $\mathrm{\mid }(a\ast {\phi }_t)(x)\mathrm{\mid }\le C_{n,s}\int \mathrm{\mid }a(y)\mathrm{\mid }\mathrm{\mid }y{\mathrm{\mid }}^{s+1}{\max }_{\mathrm{\mid }\alpha \mathrm{\mid }=s+1}{\sup }_{0\le \tau \le 1}\mathrm{\mid }{\mathrm{\partial }}^{\alpha }{\phi }_t(x-\tau y)\mathrm{\mid }dy\mathrm{.}$ Writing $q=n+s+1$, Schwartz decay says for any $r_0>q$ and $\mathrm{\mid }\alpha \mathrm{\mid }=s+1$, $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }{\phi }_t(w)\mathrm{\mid }\le C_{r_0,\alpha }{t}^{-q}(1+\mathrm{\mid }w\mathrm{\mid }\mathrm{/}t{)}^{-r_0}\le C_{r_0,\alpha }^{\mathrm{\prime }}\mathrm{\mid }w{\mathrm{\mid }}^{-q}(w\mathrm{\ne }0,t>0)\mathrm{.}$ Since $\mathrm{\mid }y\mathrm{\mid }\le \sqrt{n}$ on the support of $a$, this proves $m(x)\le C\mathrm{\mid }x{\mathrm{\mid }}^{-q}$ for $\mathrm{\mid }x\mathrm{\mid }>2\sqrt{n}$. The choice of $s$ gives $pq=p(n+s+1)>n$, so $m\in L^p$ and $a\in H^p$.

Also $m$ is positive on a nonempty open set. To see this, normalize $\mathrm{\Phi }=\phi \mathrm{/}(\int \phi )$. By Schwartz approximate identities converge in the sense of tempered distributions, ${\mathrm{\Phi }}_t\ast a\to a$ in ${\mathcal{S}}^{\mathrm{\prime }}$, so some $t_0>0$ has $a\ast {\phi }_{t_0}\equiv ̸0$; otherwise the distributional limit would be zero. This convolution is continuous, so its absolute value, and hence $m$, is positive on a nonempty open set. Thus $A:={\int }_{{\mathbb{R}}^n}m(x{)}^{p}dx=\mathrm{\parallel }a{\mathrm{\parallel }}_{H^p}^p$ is finite and positive.

For $y\in {\mathbb{R}}^n$ put $a_y(x)=a(x-y)$ and $m_y(x)=m(x-y)$. Translation invariance of the convolution and Lebesgue measure gives $M_{\phi }^0{a}_y=m_y$ and $\mathrm{\parallel }{a}_y{\mathrm{\parallel }}_{H^p}=\mathrm{\parallel }a{\mathrm{\parallel }}_{H^p}$. Pointwise sublinearity gives $M_{\phi }^0(a+a_y)\le m+m_y$, so $a+a_y\in H^p$; the reverse triangle inequality for this sublinear maximal operator gives $M_{\phi }^0(a+a_y)(x)\ge \mathrm{\mid }m(x)-m_y(x)\mathrm{\mid }\mathrm{.}$ For $Q_R=[-R,R{]}^n$, choose $R$ with ${\int }_{Q_R}{m}^p>2^{p-1}A$, possible since $2^{p-1}<1$ and $m^p\in L^1$. Take $y=T{e}_1$ with $T>2R$; then $Q_R$ and $Q_R+y$ are disjoint. The scalar inequality $\mathrm{\mid }u-v{\mathrm{\mid }}^p\ge u^p-v^p$ for $u,v\ge 0$, applied with the local copy as $u$ on each cube, gives ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }m-m_y{\mathrm{\mid }}^{p}dx\ge 2{\int }_{Q_R}{m}^{p}dx-{\int }_{Q_R}{m}_y^{p}dx-{\int }_{Q_R+y}{m}^{p}dx\mathrm{.}$ As $y\to \mathrm{\infty }$ along a coordinate ray, the last two integrals tend to zero because $m^p\in L^1$. The first term is strictly larger than $2^{p}A$. Therefore for all sufficiently large such $y$, $\mathrm{\parallel }a+a_y{\mathrm{\parallel }}_{H^p}^p\ge \int \mathrm{\mid }m-m_y{\mathrm{\mid }}^p>2^{p}A=(\mathrm{\parallel }a{\mathrm{\parallel }}_{H^p}+\mathrm{\parallel }{a}_y{\mathrm{\parallel }}_{H^p}{)}^p,$ which contradicts the ordinary triangle inequality. This proves the claimed failure within the radial-maximal definition of $H^p$.

Completeness: a complete argument is sketched here for the record. Let $(f_k)$ be Cauchy for $d$. Passing to a subsequence, assume $\mathrm{\parallel }{f}_{k+1}-f_k{\mathrm{\parallel }}_{H^p}^p\le 2^{-k}$, and set $g_k=f_{k+1}-f_k$. By Atomic characterisation of real $H^p$ for $0<p\le 1$ each $g_k$ has an atomic representation $g_k={\sum }_j{\lambda }_{k,j}{a}_{k,j}$ with ${\sum }_j\mathrm{\mid }{\lambda }_{k,j}\mathrm{\mid }\le C\mathrm{\parallel }{g}_k{\mathrm{\parallel }}_{H^p}\le C{2}^{-k\mathrm{/}p}$; the pairing bound for atoms Atoms have uniformly bounded $H^p$ quasi-norm and uniformly bounded test pairings then gives, for every $\psi \in \mathcal{S}$, $\mathrm{\mid }⟨g_k,\psi ⟩\mathrm{\mid }\le C(\psi ){\sum }_j\mathrm{\mid }{\lambda }_{k,j}\mathrm{\mid }\le C^{\mathrm{\prime }}(\psi )2^{-k\mathrm{/}p}$, so ${\sum }_k{g}_k$ defines a continuous linear functional $h$ bounded by a fixed Schwartz seminorm times ${\sum }_k{2}^{-k\mathrm{/}p}$, and converges to $h$ in ${\mathcal{S}}^{\mathrm{\prime }}$; put $f=f_0+h$. The same bound applied to the tails shows that $f_k\to f$ in ${\mathcal{S}}^{\mathrm{\prime }}$. For fixed $\phi$ and $(t,x)$, the convolutions $(f_k\ast {\phi }_t)(x)$ converge to $(f\ast {\phi }_t)(x)$; taking the supremum after pointwise convergence of each convolution gives $M_{\phi }^0(f-f_l)\le {\mathrm{lim\hspace{0.17em}inf}}_k{M}_{\phi }^0(f_k-f_l)$. Hence Fatou's lemma applied to the measurable functions $\mathrm{\mid }{M}_{\phi }^0(f_k-f_l){\mathrm{\mid }}^p$ gives $\mathrm{\parallel }f-f_l{\mathrm{\parallel }}_{H^p}^p\le {\mathrm{lim\hspace{0.17em}inf}}_k\mathrm{\parallel }{f}_k-f_l{\mathrm{\parallel }}_{H^p}^p$, which tends to $0$ as $l\to \mathrm{\infty }$; hence $f_l\to f$ in $H^p$ and $f\in H^p$. The metric $d$ is translation invariant because $M_{\phi }^0((f+h)-(g+h))=M_{\phi }^0(f-g)$. No Banach duality is used in this argument, and the completion obtained is the space $H^p$ itself.