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From an Lp norm Cauchy sequence choose a subsequence with successive difference norms at most $2^{-j}$. Minkowski inequality and monotone convergence theorem make the sum of the absolute differences an $L^p$ function, hence finite almost everywhere. The telescoping series defines an $L^p$ limit; the geometric bound on its tail proves norm convergence of the subsequence, and the original Cauchy property proves convergence of the whole sequence. No sigma-finiteness assumption is needed.

For $p\ge 2$ and $p^{\prime }=p/(p-1)$, interpolate the two-coordinate map $(x,y)\mapsto (x+y,x-y)$ between ${\ell }^1\to {\ell }^{\infty }$ norm one and ${\ell }^2\to {\ell }^2$ norm $\sqrt{2}$. This proves the scalar inequality with factor $2^{1/p}$. Integrating that estimate and applying Minkowski inequality to $|f{|}^{p^{\prime }}+|g{|}^{p^{\prime }}$ in $L^{p/p^{\prime }}$ gives the displayed function inequality.

For $1<p<\infty$ and nonzero elements $f,g$ of the same real or complex Lp space, equality in the Minkowski inequality holds if and only ifEquivalently, $g=cf$ almost everywhere for a single positive real number $c$. The multiplier must be constant across $X$; matching signs at each input alone is insufficient. For complex numbers, an arbitrary complex scalar multiple is also insufficient: $g=if$ gives $\Vert f+g{\Vert }_p=\sqrt{2}\Vert f{\Vert }_p<2\Vert f{\Vert }_p$ for nonzero $f$. If either element is zero in Lp space, equality holds automatically, even if a representative has nonzero values on a null set.

The reason is strictly convex normed space geometry. Normalize $f,g$ to unit Lp norm and use weights $\Vert f{\Vert }_p/(\Vert f{\Vert }_p+\Vert g{\Vert }_p)$ and $\Vert g{\Vert }_p/(\Vert f{\Vert }_p+\Vert g{\Vert }_p)$. Distinct normalized functions give a convex combination of Lp norm strictly below one. This follows by integrating the strict convexity of $z\mapsto |z{|}^p$ on the real numbers or complex numbers; equality therefore forces the normalized functions to agree almost everywhere.

At $p=1$, equality instead means $|f+g|=|f|+|g|$ almost everywhere: the nonnegative difference has zero integral. For real functions, this is $fg\ge 0$ almost everywhere. For complex functions, their nonzero values must have the same argument wherever both are nonzero, but their magnitude ratio may vary. The disjoint-coordinate vectors $(1,0)$ and $(0,1)$ illustrate equality without constant proportionality.

At $p=\infty$, no such constant proportionality condition is necessary: $(1,0)$ and $(1,1)$ have L-infinity norms one and a sum of L-infinity norm two. Nor must an essential supremum be attained: under Lebesgue measure on $(0,1)$, $f(x)=g(x)=x$ give equality $2=1+1$, although neither function takes the value one. The endpoint Lp spaces need not be strictly convex normed spaces.

For $0<p<1$, write $Q_p(f)=(\int |f{|}^{p}d\mu {)}^{1/p}$ to emphasize that this is a quasi-norm, as in the Lq quasi-norm branch. On two coordinates with counting measure, $u=(1,0)$ and $v=(0,1)$ give $Q_p(u)=Q_p(v)=1$, but $Q_p(u+v)=2^{1/p}>2$. At $p=1/2$, this is the explicit failure $4>1+1$ in the right panel of the diagram. The midpoint $(1/2,1/2)$ has $Q_p=2^{1/p-1}>1$, so the corresponding unit sublevel set is not a convex set.

A useful replacement survives: for nonnegative real numbers $a,b$, $(a+b{)}^p\le a^p+b^p$. Together with the scalar triangle inequality, this yieldsConsequently $d_p(f,g)=Q_p(f-g{)}^p$ is a metric on the equivalence classes of measurable functions with finite $Q_p$, even though $Q_p(f-g)$ itself need not be a metric. Do not replace the failed Minkowski inequality by a universal reversed bound for signed or complex functions: $g=-f\ne 0$ would give a zero sum but two positive separate sizes.

For a nonnegative symmetric unit-integral kernel for density estimation and $f\in H^2(\mathbb{R})$, the integral Taylor remainder and Minkowski inequality give $\Vert K_h\ast f-f{\Vert }_2\le h^2{\mu }_2(K)\Vert f^{\prime \prime }{\Vert }_2/2$. Translation preserves the L2 norm. For a signed kernel the same proof requires the absolute moment $\int u^2|K(u)|du$.

For measurable functions $f,g$ with values in the real numbers or complex numbers on the same measure space $(X,\mathcal{A},\mu )$, and $1\le p\le \infty$, membership in the same Lp space givesThus adding two functions cannot make their combined Lp norm exceed the sum of their separate Lp norms. The measure need not be finite or a probability measure. The hypotheses require finite Lp norms; $p=\infty$ uses the essential supremum of the absolute value, not an ordinary pointwise supremum. For complex numbers, the absolute value is the complex modulus. As usual, Lp spaces identify functions equal almost everywhere. ^[ref]

The Minkowski inequality makes the Lp norm a workable notion of size: it supplies the triangle inequality needed to turn $d_p(f,g)=\Vert f-g{\Vert }_p$ into a metric on Lp space. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate function $h$ cannot shorten the direct distance measured by this metric. Applied to two error functions, it gives a bound on the total error even when the errors reinforce one another; Minkowski inequality for error aggregation gives explicit examples.

The geometry is already visible in the Euclidean plane. Under counting measure on two coordinates, the closed unit balls for the L1 norm, Euclidean norm, and L-infinity norm are respectively $|x|+|y|\le 1$, $x^2+y^2\le 1$, and $\max (|x|,|y|)\le 1$. Every one is a convex set: if $u,v$ have Lp norm at most one, the Minkowski inequality and absolute homogeneity of a norm give $\Vert tu+(1-t)v{\Vert }_p\le t+(1-t)=1$ for $0\le t\le 1$. The Euclidean norm has no flat boundary segments, whereas the other two norms do. This is the distinction captured by a strictly convex normed space and developed in equality in Minkowski inequality.

A short Hölder's inequality proof outline handles $1<p<\infty$. First $|f+g{|}^p\le 2^{p-1}(|f{|}^p+|g{|}^p)$, so $f+g$ has finite Lp norm before any division. With conjugate exponents $p$ and $q=p/(p-1)$, set $h=|f+g{|}^{p-1}$; then $\Vert h{\Vert }_q=\Vert f+g{\Vert }_p^{p-1}$. The scalar triangle inequality followed by Hölder's inequality givesDivide if $\Vert f+g{\Vert }_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the result by the essential supremum. The same scalar triangle inequality works for real numbers and complex numbers.

Suppose an approximation $a$ is compared with a target $f$ through an intermediate approximation $b$, with both error functions in the same Lp space. The identity $f-a=(f-b)+(b-a)$ and the Minkowski inequality combine two error budgets without any independence assumption. The L1 norm controls aggregate absolute error, the L2 norm controls the square root of aggregate quadratic loss, and the L-infinity norm controls the largest error almost everywhere. For equally weighted samples with the probability measure $1/n$ on each coordinate, the L2 norm is the root mean square error; the inequality still applies when this normalization is used consistently.

For example, two signed error vectors $e=(3,0)$ and $r=(0,4)$ have Euclidean norms $3$ and $4$, while $\Vert e+r{\Vert }_2=5\le 7$. Reinforcing errors $e=(3,0)$ and $r=(4,0)$ attain the bound $7$; opposing errors $e=(3,0)$ and $r=(-3,0)$ cancel completely. The Minkowski inequality always allows reinforcement. A smaller combined budget requires additional information about the errors, not just their separate norms.

Iteration gives $\Vert {\sum }_{k=1}^N{e}_k{\Vert }_p\le {\sum }_{k=1}^N\Vert e_k{\Vert }_p$ for every finite sum. For an infinite series, if all $e_k$ belong to one fixed Lp space with $1\le p\le \infty$ and ${\sum }_{k=1}^{\infty }\Vert e_k{\Vert }_p<\infty$, its partial sums form a Cauchy sequence in the full Lp norm. Completeness of this Banach space then gives norm convergence to an element of the same Lp space. A finite sum bound alone does not establish convergence, and completeness is an additional ingredient beyond the Minkowski inequality. ^[ref]

For vectors $u,v\in {\mathbb{R}}^n$ or ${\mathbb{C}}^n$, counting measure turns the Minkowski inequality intoFor $p=\infty$, replace each displayed norm by the maximum of its coordinate absolute values. Take $u=(1,0)$ and $v=(0,1)$. Both vectors have Lp norm one for every $p\ge 1$, but their sum has L1 norm $2$, Euclidean norm $\sqrt{2}$, and L-infinity norm $1$. Thus the same vector addition gives equality $2=1+1$ at $p=1$, strict inequality $\sqrt{2}<1+1$ at $p=2$, and strict inequality $1<1+1$ at $p=\infty$.

By contrast, $u=(1,1)$ and $v=2u=(2,2)$ give equality for every $p\ge 1$: absolute homogeneity of a norm yields $\Vert u+v{\Vert }_p=3\Vert u{\Vert }_p=\Vert u{\Vert }_p+\Vert v{\Vert }_p$. At $p=2$, the three sizes are $\sqrt{2}$, $2\sqrt{2}$, and $3\sqrt{2}$. Opposite vectors behave very differently: $u+(-u)=0$, so the inequality is strict whenever $u$ is nonzero. Cancellation is allowed; the bound describes the largest possible total size.

This is the integral version of the Minkowski inequality.

For suitable measurable $F(x,y)$ and $1\le p<\infty$,Pair the left side with a unit vector in the dual $L^q$ space, use Tonelli's theorem, and apply Hölder's inequality in $y$.

Let $(f_n)$ be a Cauchy sequence in the L2 space $L^2(\mathbb{R})$, choosing measurable representatives. There is a subsequence $(f_{n_k})$ withWrite $d_k=f_{n_{k+1}}-f_{n_k}$ and $S_N={\sum }_{k=1}^N|d_k|$. Repeated Minkowski inequality gives $\Vert S_N{\Vert }_2\le {\sum }_{k=1}^N{2}^{-k}\le 1$. As $S_N^2$ increases pointwise, the monotone convergence theorem givesHence $S={\sum }_k|d_k|$ is finite almost everywhere and represents an L2 function. Definewhere the series converges, and set $f=0$ on the exceptional Lebesgue null set. It is a measurable function, and $|f|\le |f_{n_1}|+S$ almost everywhere, so Minkowski inequality gives $f\in L^2(\mathbb{R})$.

For fixed $k$, Fatou's lemma applied to squares of finite tails and then Minkowski inequality giveThus the subsequence converges to $f$ in norm. Given $\epsilon >0$, choose a sufficiently late $n_k$ with $\Vert f-f_{n_k}{\Vert }_2<\epsilon /2$ and with $\Vert f_n-f_{n_k}{\Vert }_2<\epsilon /2$ for every sufficiently large $n$, using the original Cauchy sequence property. The triangle inequality then gives $\Vert f_n-f{\Vert }_2<\epsilon$. ThereforeThis proves the needed instance of completeness of Lp spaces without assuming the result being proved.

Use Lebesgue measure restricted to the Lebesgue measurable set $E$. For $1\le p<\infty$, start with the measurable real-valued functions $f$ on $E$ satisfying ${\int }_E|f{|}^{p}dm<\infty$. Define an equivalence relation by $f\sim g$ when $f=g$ almost everywhere, and let $L^p(E)$ consist of their equivalence classes. Addition and scalar multiplication are defined using representatives. The Lp norm isFor $p=\infty$, use measurable functions bounded almost everywhere, apply the same identification, and define the L-infinity norm byThese expressions are independent of the representative. For finite $p$, zero norm is equivalent to $f=0$ almost everywhere: each set $\{|f|>1/n\}$ must have measure zero, and their union is $\{f\ne 0\}$. The same implication follows from the essential supremum for $p=\infty$. Absolute homogeneity is immediate; Minkowski inequality proves the triangle inequality and closure under addition. Thus these are normed vector spaces. Using complex-valued representatives gives their complex versions by the same construction.

Minkowski inequality states that on a measure space $(X,\Sigma ,\mu )$, for $1\le p\le \infty$ and measurable real or complex $f,g$ of finite Lp norm,For $p=1$, integrate the scalar triangle inequality $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the same result by taking the essential supremum.

For $1<p<\infty$, first ensure that the left side is finite. The convex function $s\mapsto s^p$ on $[0,\infty )$ givesso $f+g$ has finite Lp norm. Set $q=p/(p-1)$ and $H=\Vert f+g{\Vert }_p$. Since $(p-1)q=p$, Holder inequality givesIf $H=0$ the result is immediate; otherwise divide by $H^{p-1}$. This proves the triangle inequality for every claimed exponent.

Consider the complex-linear two-coordinate map $H(x,y)=(x+y,x-y)$. The triangle inequality shows that its operator norm from ${\ell }^1$ to ${\ell }^{\infty }$ is one; equality is attained at $(1,0)$. The complex parallelogram identityshows that its norm from ${\ell }^2$ to ${\ell }^2$ is $\sqrt{2}$. Interpolate these endpoint bounds by the Riesz-Thorin theorem with parameter $\eta =2/p$. ThenDividing the output norm by $2^{1/p}$ proves the scalar estimate:Apply it pointwise to $f,g$ and integrate its $p$th power. With $r=p/p^{\prime }=p-1>1$, this givesBy Minkowski inequality, the expression on the right is at mostTherefore the requested function inequality holds, namely the dual-exponent Clarkson inequality.

Completeness means that every norm-Cauchy sequence has a limit belonging to the same Lp space, with $\Vert f_n-f{\Vert }_p\to 0$. To prove it, choose a subsequence $f_{n_j}$ with $\Vert f_{n_{j+1}}-f_{n_j}{\Vert }_p\le 2^{-j}$. The functionshave $p$-norm bounded by $\Vert f_{n_1}{\Vert }_p+1$, by Minkowski inequality. They increase to $H$, and Fatou's lemma gives $\int H^p\le (\Vert f_{n_1}{\Vert }_p+1{)}^p$. Thus $H$ is finite almost everywhere, so the subsequence telescopes absolutely there to a measurable limit $f$ with $|f|\le H$ and $f\in L^p$.

The same argument on the tails gives $\Vert f-f_{n_j}{\Vert }_p\le {\sum }_{k\ge j}{2}^{-k}\to 0$. Given $ϵ$, choose such a subsequence term beyond the Cauchy threshold and use $\Vert f_n-f{\Vert }_p\le \Vert f_n-f_{n_j}{\Vert }_p+\Vert f_{n_j}-f{\Vert }_p$. This proves convergence of the full sequence and completeness for every $1\le p<\infty$, on an arbitrary measure space.

The Minkowski inequality states the triangle inequality for every $L^p$ norm with $1\le p\le \infty$. Apply it to $f_n=(f_n-f)+f$ and then to $f=(f-f_n)+f_n$. The two bounds combine toThus norm convergence implies convergence of the norms themselves.

The real L2 space on a probability space consists of equivalence classes of measurable functions $X:\Omega \to \mathbb{R}$ satisfying $\mathbb{E}[X^2]<\infty$, where two random variables represent the same element if they are equal almost surely. Its norm isPassing to equivalence classes makes this a genuine norm: $\Vert X{\Vert }_2=0$ exactly when $X=0$ almost surely. The Minkowski inequality gives the triangle inequality.

Let $(X_n)$ be a Cauchy sequence in this norm. Choose an increasing sequence of indices $n_k$ such thatThis is possible by choosing $n_k$ beyond a Cauchy threshold for tolerance $2^{-k}$. Set $D_k=X_{n_{k+1}}-X_{n_k}$. The Cauchy-Schwarz inequality on a probability space gives $\mathbb{E}|D_k|\le \Vert D_k{\Vert }_2$. By Tonelli theorem,Thus the series of absolute differences is finite almost surely. Define $X=X_{n_1}+{\sum }_{k\ge 1}{D}_k$ on this probability-one event, and give it the value zero on its complement. It is a measurable function, and $X_{n_k}\to X$ almost surely.

For $l>k$, the Minkowski inequality givesApply the Fatou lemma to the squared difference as $l\to \infty$. It follows thatIn particular $X-X_{n_1}\in L^2$, and therefore $X\in L^2$ by the Minkowski inequality. This proves convergence in L2 of the subsequence. To recover the entire sequence, for a prescribed $\epsilon >0$ choose a Cauchy threshold $N$ for $\epsilon /2$, then choose $k$ with $n_k\ge N$ and $\Vert X-X_{n_k}{\Vert }_2<\epsilon /2$. For every $n\ge N$,Hence $\boxed{X_n\to X\text{ in }{L}^2}$, proving completeness of Lp spaces for $p=2$ without assuming the desired completeness in the argument.

Concavity of $\log$ gives $\log (a/p+b/q)\ge p^{-1}\log a+q^{-1}\log b$ because $1/p+1/q=1$. Exponentiating proves Young inequalitywith equality precisely when $a=b$. Equivalently it follows from convexity of the exponential.

Define ${\ell }^p=\{x=(x_j):{\sum }_j|x_j{|}^p<\infty \}$ with $\Vert x{\Vert }_p=({\sum }_j|x_j{|}^p{)}^{1/p}$. Positivity, definiteness and homogeneity are immediate. Summing Young inequality for normalized sequences gives Hölder's inequality $\sum |x_j{y}_j|\le \Vert x{\Vert }_p\Vert y{\Vert }_q$. For a finite truncation $z=x+y$, it yieldsDividing and passing to larger truncations proves Minkowski inequality and shows $x+y\in {\ell }^p$, so this is a normed vector space.

Let $x^{(m)}$ be a norm-Cauchy sequence. Each coordinate is Cauchy because $|x_j^{(m)}-x_j^{(l)}|\le \Vert x^{(m)}-x^{(l)}{\Vert }_p$, and let its limit be $x_j$. The sequence norms are bounded by some $C$. For every finite $N$, passage to coordinate limits gives ${\sum }_{j\le N}|x_j{|}^p\le C^p$, so $x\in {\ell }^p$. Given $\epsilon >0$, choose $M$ with $\Vert x^{(m)}-x^{(l)}{\Vert }_p<\epsilon$ for $m,l\ge M$. Letting $l\to \infty$ in finite truncations and then $N\to \infty$ gives $\Vert x^{(m)}-x{\Vert }_p\le \epsilon$ for $m\ge M$. Thus ${\ell }^p$ is complete, hence a Banach space.

Apply Hölder's inequality to the finite sum with the second sequence $|x_j+y_j{|}^{p-1}$:Divide by the last power if the sum is nonzero. The resulting bound is independent of $N$, so passing to the limit proves both membership in ${\ell }^p$ and the Minkowski inequality

The Lp space is the vector space of equivalence classes, modulo equality almost everywhere, of measurable functions $f$ with $\int |f{|}^{p}d\mu <\infty$, equipped with $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. To prove it is a Banach space, let $(f_n)$ be Cauchy in this norm. Choose a subsequence $(f_{n_j})$ such thatTake measurable representatives and put $g_j=|f_{n_{j+1}}-f_{n_j}|$, $G_M={\sum }_{j=1}^M{g}_j$. The Minkowski inequality gives $\Vert G_M{\Vert }_p\le {\sum }_j{2}^{-j}\le 1$. By the monotone convergence theorem, $G={\sum }_{j\ge 1}{g}_j$ satisfies $\int G^{p}d\mu \le 1$. Consequently $G$ is finite almost everywhere, and the telescoping seriesconverges absolutely almost everywhere. Define it arbitrarily on the null exceptional set. The bound $|f|\le |f_{n_1}|+G$ puts $f$ in $L^p$.

For the tails, monotone convergence and Minkowski again giveSince the original sequence is Cauchy, the triangle inequality then gives $f_n\to f$ in norm, not merely along the subsequence. HenceThis proof needs no finiteness or sigma-finiteness assumption on the measure space.

Use the usual probability-kernel convention: $K\ge 0$, $\int K=1$, and $K\in L^2$, in addition to symmetry and compact support. These conditions are needed for the stated finite bound; symmetry and compact support by themselves do not impose normalization or nonnegativity. Interpret the differentiability assumption in the usual Sobolev space sense $f\in H^2(\mathbb{R})$, or assume classical regularity sufficient for the integral Taylor remainder. The kernel density estimator isThe bias-variance decomposition of mean squared error, followed by Tonelli theorem, yieldsIndependence and a change of variables give the exact integrated variance of a kernel density estimatorAll changes of integration order are justified either by nonnegativity or by the displayed finite integrals.

For the bias of a kernel density estimator, symmetry gives $\int uK(u)du=0$. Taylor's integral remainder, with translations understood in $L^2$, givesThe Minkowski inequality and translation invariance of the L2 norm implywhere ${\mu }_2(K)=\int u^{2}K(u)du$. Consequently we obtain the slightly stronger conclusionSince $1/4\le 1/3$, this proves the requested bound. For signed kernels, the same argument has $\int u^2|K(u)|du$ in place of ${\mu }_2(K)$; the printed bound is not generally justified merely by signed-kernel symmetry. For example, let $U_a=(2a{)}^{-1}{1}_{[-a,a]}$ and $K=(4/3)U_1-(1/3)U_2$. This symmetric compactly supported signed kernel has integral one and ${\mu }_2(K)=0$, but its fourth moment is nonzero. For a normal density its convolution has nonzero bias at any fixed positive bandwidth. Sending $n\to \infty$ would therefore contradict a bound containing only the integrated variance term.

Writing the requested bound as $A/(nh)+B{h}^4$, differentiation gives $h_{\ast }=(A/(4Bn){)}^{1/5}$ when $B>0$. Thus $h\asymp n^{-1/5}$ gives mean integrated squared error of order $n^{-4/5}$. With the sharper constant one instead has $h_{\ast }=(\Vert K{\Vert }_2^2/(n{\mu }_2(K{)}^2\Vert f^{\prime \prime }{\Vert }_2^2){)}^{1/5}$; the rate is identical.

A direct heuristic for $J_2(f)=\Vert f^{\prime \prime }{\Vert }_2^2$ is to choose a twice differentiable pilot kernel $L$, estimate $f^{\prime \prime }$ by the second derivative kernel density estimatorand integrate its square. Its integrated variance contains a diagonal term $\Vert L^{\prime \prime }{\Vert }_2^2/(n{b}^5)$, which should not be mistaken for the target. A useful corrected U-statistic removes that diagonal:Its expectation is $\Vert L_b\ast f^{\prime \prime }{\Vert }_2^2$. Thus smoothing bias of an estimator, rather than the diagonal contribution, remains. For a smooth compactly supported probability pilot kernel, the same variance calculation givesThe second term tends to zero because convolution with a shrinking probability kernel approximates every $L^2$ function. Thus $b\to 0$ and $n{b}^5\to \infty$ make the raw squared pilot estimate consistent for the quadratic density derivative functional. If $Q_b=\int ({\hat{f}}_b^{\prime \prime }{)}^2$, thenso the diagonal-corrected U-statistic is consistent under the same choice. Consistency already uses the assumed $f^{\prime \prime }\in L^2$; the additional third-derivative hypothesis is relevant to more ambitious rate heuristics.

Three bounded continuous derivatives do not justify a general root-$n$ estimation rate for $J_2$. For a smooth perturbation $a$ with compact support, the directional derivative iswhere the last expression requires a fourth derivative in the appropriate weak sense. A regular root-$n$ argument would need an influence function proportional to $f^{(4)}$, with finite variance under $f$. Three derivatives do not supply that condition.

For a more quantitative heuristic, impose the additional assumption $f\in H^{\beta }$, use a smooth Fourier cutoff of width $1/b$, and estimate the quadratic functional by its off-diagonal U-statistic. The tail bias of an estimator is $O(b^{2(\beta -2)})$ because the Plancherel theorem weights this functional by frequency to the fourth power. The degenerate part of its standard deviation is $O(n^{-1}{b}^{-9/2})$: the squared $L^2$ norm of the quadratic kernel scales as $b^{-9}$. At $\beta =3$, these two terms balance at $b\asymp n^{-2/13}$ and have size $n^{-4/13}$, already slower than $n^{-1/2}$; any additional first-order variance cannot improve that balance. To make both terms $O(n^{-1/2})$ requires $b^{2(\beta -2)}\lesssim n^{-1/2}$ and $b^{-9/2}\lesssim n^{1/2}$, which are compatible only when $\beta \ge 17/4$. This is an illustrative smoothness calculation under extra Sobolev and moment assumptions, not a rate theorem asserted solely from bounded third derivatives. Particular much smoother probability density functions may of course permit root-$n$ estimation.