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An $\epsilon$-net of a metric space is a set $G$ such that every point lies within distance $\epsilon$ of a point of $G$. A Cartesian grid of coordinate spacing $\eta$ is a finite $\sqrt{d}\eta$-net of a bounded cube in the Euclidean norm. A Lipschitz bound transfers control on the grid to control over the whole cube.

A function is globally Lipschitz when one Lipschitz bound holds for every pair of points in its domain. Unlike local Lipschitz continuity, this condition gives a uniform growth control over the whole domain.

For a strictly positive probability density function with Lipschitz bound $L$, $p_x(r)={\int }_{x-r}^{x+r}f(y)dy$ satisfies $|p_x(r)-2rf(x)|\le L{r}^2$. If $0<s<1$ and $s\le f(x{)}^2/L$, monotonicity gives $p_x^{-1}(s)\le s/f(x)$ and hence $|2f(x)p_x^{-1}(s)-s|\le L{s}^2/f(x{)}^2$. The Lipschitz density height bound prevents extending this local argument to every $s\in (0,1)$ by assuming $f(x)\ge \sqrt{L}$ for a strictly positive density.

The least $L\ge 0$ for which $d(f(x),f(y))\le Ld(x,y)$ for every $x,y$ is the Lipschitz constant of $f$. Any admissible $L$ is a Lipschitz bound.

For a probability density function $f$ on $\mathbb{R}$ with Lipschitz bound $L>0$, the triangular lower envelope $f(x+u)\ge (f(x)-L|u|{)}_{+}$ gives $f(x{)}^2\le L$. If $f$ is strictly positive everywhere, the inequality is strict. Equality forces the entire probability density function to coincide with the triangular envelope, which vanishes outside a finite interval.

A function is locally Lipschitz when every point has a neighborhood on which the function satisfies a Lipschitz bound. Continuously differentiable functions are locally Lipschitz.

The empirical spectral measures of the two real symmetric matrices arewhere ${\delta }_t$ is the Dirac measure at $t$. Thus $⟨L_N,f⟩=N^{-1}{\sum }_{i}f({\lambda }_i)$. Pair the eigenvalues in the specified increasing order and apply the triangle inequality followed by Lipschitz continuity with bound $1$:This deterministic estimate uses only the Lipschitz bound; no entry independence or distributional hypothesis is needed here.

A function $f:\mathbb{R}\to \mathbb{R}$ satisfies Lipschitz continuity with bound $L\ge 0$ ifHere $L$ is an admissible Lipschitz bound; the least admissible value is the Lipschitz constant. The later parts use the admissible bound $1$.

Use the fixed common entry law from the printed setup: $Y=\sqrt{N}{X}_{ij}$ has mean zero and second moment $1$ for each upper-triangular entry, including the diagonal. LetSince $\mathbb{E}Y=0$, the removed part after centered truncation of a Wigner matrix isConsequently $\mathbb{E}{\Delta }_{ij}^2=N^{-1}\mathrm{Var}(Z_C)\le N^{-1}t(C)$. Symmetry gives $\mathrm{Tr}{\Delta }^2={\sum }_{i,j}{\Delta }_{ij}^2$. There are $N$ diagonal terms and $N(N-1)$ off-diagonal terms in this sum, soMirrored entries are counted twice, as they must be; independence of those mirrored entries is neither true nor needed for this expectation calculation.

By the preceding bound and the Markov inequality,The dominated convergence theorem applies to $Y^2{1}_{\{|Y|\ge C\}}\le Y^2$, so $t(C)\to 0$. Choose $C$ with $t(C)<{\epsilon }^3$. ThenThis second moment bound for spectral truncation depends only on $\epsilon$ and the common law of $Y$, and is uniform over functions with Lipschitz bound $1$. No fourth-moment hypothesis is required. If the scaled entry law were allowed to vary arbitrarily with $N$, a uniform choice would instead require uniform decay of its second-moment tails; the intended fixed-law setting supplies exactly that condition.

The stated bound is Lipschitz continuity with constant $1$. For every $x$ and every $\epsilon >0$, taking $\delta =\epsilon$ ensures $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. The same $\delta$ works at every point, so $f$ is both continuous and uniformly continuous.

For an integer $N\ge 1$, partition $[0,1]$ into $N$ equal intervals and define a step functionAt the remaining endpoint set $g_N(1)=f((N-1)/N)$, so the final interval may be taken closed on the right. The Lipschitz bound givesChoosing $N\ge 1/\epsilon$ proves the requested uniform step approximation on a compact interval:A sufficiently fine partition gives a piecewise constant approximation uniformly over the whole interval.

The continuous function $f$ and every finite step function $g_N$ are Riemann integrable. Put $c_j=f(j/N)$. Integrating each constant piece explicitly givesSince the sine differences have absolute value at most $2$, the triangle inequality and uniform approximation giveTo prove convergence, fix $\epsilon >0$ and first choose $N$ with $1/N<\epsilon /2$. With this $N$ held fixed, the finite number ${\sum }_j|c_j|$ is independent of $n$, so the second term is also less than $\epsilon /2$ for sufficiently large $n$. ThereforeFirst make the approximation error small, then let the oscillation frequency grow. This is the step-function proof of the Riemann-Lebesgue lemma; it does not require assuming that a Lipschitz function has an everywhere-defined derivative.

Let $m=\mathbb{E}\Vert X_1{\Vert }_2$ and $q=\sqrt{n}K\ge 2$. The Lipschitz continuity of cosine and the Cauchy-Schwarz inequality implyThe centered expectation term contributes $nm$; omitting it would leave the Lipschitz bound incomplete. In particular $S_n$ has continuous sample paths on the compact cube, so its maximum exists and is measurable.

PutChoose a Cartesian finite net $G\subset [-K,K{]}^d$ whose coordinate spacings are at most $\eta$, including the endpoints. Every point of the cube is within Euclidean norm distance $\sqrt{d}\eta$ of $G$, andOn the event $L_n\le n{q}^a$, replacing any $u$ by a nearby grid point changes $S_n$ by at most $\sqrt{dn}$. Since $R>8\sqrt{d}$ and $\log (n{K}^2)\ge \log 4$, this is less than $\frac{R}{4}\sqrt{n\log (n{K}^2)}$. ThereforeThe Markov inequality bounds the first event by $2m{q}^{-a}$. At a fixed $v$, the summands are centered, independent and identically distributed, and have absolute value at most $2$. The supplied Hoeffding inequality, with $M=2$, bounds each grid event byA union bound now givesThe choice of $a$ balances the cost of a large random Lipschitz constant against the covering number of the cube: $d(1+a)-R^2/64=-a$. Consequently $C=2m+2\cdot 4^d$ works, withThis uses only the first moment $m$ and has a constant independent of $n,K,R$. The finite net is deterministic even though the Lipschitz constant is random.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

For $M=\max \{|s|,|t|,1\}$, factorization of a difference of powers givesConsequently the polynomial $f$ obeys the Lipschitz bound

Put $G={\max }_i\Vert a_i{\Vert }_2$. By the Cauchy-Schwarz inequality,Interchanging $x$ and $y$ proves the Lipschitz bound

The subgradient method chooses $g_k\in \partial f(x_k)$ and a step size $t_k>0$, then setsAssume, as the question's use of $\min f$ requires, that a minimizer $x_{\ast }$ exists, and write $R=\Vert x_0-x_{\ast }{\Vert }_2$. Since every subgradient here has Euclidean norm at most $G$, the standard best-iterate estimate isTaking a suitable constant step when the target accuracy is known, or a standard diminishing sequence, gives error at most $ϵ$ initerations, so the requested exponent is $p=2$.

At an interior target point in regular fixed-design nonparametric regression, mean functions with Lipschitz bound $L$ have pointwise minimax risk of order $(L{\sigma }^2/n{)}^{2/3}$ for fixed positive $L,\sigma$ as $n\to \infty$. A smoothing bandwidth of order $({\sigma }^2/(n{L}^2){)}^{1/3}$ balances squared bias of an estimator and variance; the two-point lower bound with a triangular bump gives the matching lower rate.

For equally sized Hermitian matrices and a real test function with Lipschitz bound one, the difference of its averages under their empirical spectral measures is at most $N^{-1/2}$ times the Frobenius norm of their difference. Pair the increasing eigenvalues, apply the triangle inequality and Cauchy-Schwarz inequality, then use the Hoffman–Wielandt inequality.

For fixed-design nonparametric regression, compare $m_0=0$ with $m_1(x)=L(h-|x-x_0|{)}_{+}$. This pair has parameter separation $Lh$ at $x_0$, Lipschitz bound $L$, and Kullback-Leibler divergence at most $L^2{h}^2(2nh+1)/(2{\sigma }^2)$. Choosing $h\asymp ({\sigma }^2/(n{L}^2){)}^{1/3}$ controls total variation distance by Pinsker's inequality and yields the pointwise minimax rate for Lipschitz regression through the Le Cam two-point lemma.

A uniformly continuous real function on a closed bounded interval can be approximated uniformly by finite step functions. Given $\epsilon >0$, choose a mesh smaller than the uniform-continuity distance and set the approximant equal to a sampled value on each interval. Then the supremum norm of the approximation error is at most $\epsilon$. If the function has a Lipschitz bound $L$, a mesh of maximum length $h$ gives the explicit error bound $\Vert f-g{\Vert }_{\infty }\le Lh$.