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Let $\delta =\hat{\beta }-{\beta }^0$. Standard sub-Gaussian random variable and Gaussian-product concentration gives, for $n>\log p$,Thus $A=4$ makes this probability at most $2/p$. On the complementary event, the basic inequality and Holder inequality giveThe parenthesis is at most $2\Vert \delta {\Vert }_1$ by the triangle inequality and at most $2\Vert {\beta }^0{\Vert }_1$ by $\Vert \delta {\Vert }_1\le \Vert \hat{\beta }{\Vert }_1+\Vert {\beta }^0{\Vert }_1$. Substituting $\lambda =A\sigma v\sqrt{\log p/n}$ proves ${\Omega }_1$, whose probability tends to one.

By the Holder inequality with exponents $3,2,6$,Part ii and the Sobolev embedding $H^1\hookrightarrow L^6$ yieldEquivalently,

For smooth periodic fields, the Holder inequality and $H^1\hookrightarrow L^6$ giveSimilarly,Adding the bounds proves that the skew-symmetrized transport form extends continuously from smooth fields to ${\tilde{V}}^3$ andThe periodic Poincare inequality givesHence

With the usual objectivethe Karush-Kuhn-Tucker conditions for the Lasso areComparing the objective at ${\hat{\beta }}_{\lambda }$ and ${\beta }^0$, expanding the square, and cancelling the noise norm gives exactly

Choose $\lambda =A\sigma \sqrt{\log p/n}$. Since every column has norm $\sqrt{n}$, each $X_j^T\epsilon /n$ is sub-Gaussian with parameter $\sigma /\sqrt{n}$. A union bound givesOn the complementary event, Holder inequality and the triangle inequality bound the preceding right-hand side bywhich proves the stated prediction bound.

For the Dantzig selector, use the constraint $n^{-1}\Vert X^T(Y-X\beta ){\Vert }_{\infty }\le \lambda$. The Lasso KKT conditions make ${\hat{\beta }}_{\lambda }$ feasible, so a Dantzig minimizer $\tilde{\beta }$ has no larger ${\ell }^1$ norm. If $\tilde{\beta }$ were not the Lasso solution, uniqueness from invertibility of $X^{T}X$ would imply that some active coordinate $j$ has strict signed KKT slack. For small $\eta >0$, setOnly score coordinate $j$ changes, toward its feasible boundary, so ${\beta }_{\eta }$ remains feasible. For small $\eta$, signs on the active coordinates do not change, andby strict diagonal dominance. This contradicts Dantzig optimality, so the two estimators coincide.

A kernel for density estimation is a bounded nonnegative integrable function $K$ with ${\int }_{\mathbb{R}}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator is

Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,The Cauchy-Schwarz inequality and variance additivity for independent random variables giveNonnegativity also gives $\mathbb{E}|V_1-\mathbb{E}{V}_1|\le 2\mathbb{E}{V}_1\le 4\Vert K{\Vert }_{\infty }{f}_h(x)$. Taking the better estimate at each $x$ and applying Tonelli theorem proves

For nonnegative $a,b$ and $0\le \theta \le 1$, $\min a,b\le a^{\theta }{b}^{1-\theta }$. Taking $\theta =2\rho$ yieldsFor $\rho >0$, the Holder inequality with conjugate exponents $(1-\rho {)}^{-1}$ and ${\rho }^{-1}$ givesIf $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has probability density function $f_h$. Sincethe last integral is at most $(1+h{)}^{\delta }\int (1+|x|{)}^{\delta }f(x)dx$. Substitution proves the second displayed bound. The case $\rho =0$ is the first bound integrated using $\int f_h=1$ and follows directly.

Let $q=2p+2$ and multiply the vorticity equation by ${\omega }^{q-1}$, first for an odd power as in the hint. Periodic integration by parts and $\nabla \cdot u=0$ giveThe diffusion term is nonnegative, while the Holder inequality bounds the right-hand side by $\Vert g{\Vert }_q\Vert \omega {\Vert }_q^{q-1}$. ConsequentlyOn the finite-volume torus, $L^q$ norms increase to the essential supremum as $q\to \infty$. Hence the damped-vorticity maximum estimate gives

For $Y=X{\beta }^0+\eta$, the Lasso objective $\Vert Y-X\beta {\Vert }_2^2/(2n)+\lambda \Vert \beta {\Vert }_1$ has this prediction bound whenever $\Vert X^T\eta {\Vert }_{\infty }/n\le \lambda$. Compare its value at $\hat{\beta }$ and ${\beta }^0$, apply Holder inequality to the score, and use the triangle inequality. No restricted eigenvalue condition is required.

For a convergent continuous local martingale and $r>1$, factor $\mathcal{E}(X)$ into $\mathcal{E}(rX{)}^{1/r^2}$ and an ordinary exponential. Apply the Holder inequality and then the concave Jensen inequality to the fractional power $2/(r+1)$. The displayed lower bound is informative when the terminal exponential moment is finite.

Expand moments of the inner sum and group equal power-sum differences. Their multiplicities are bounded by $J_{k,r}(Z)$ through the Cauchy-Schwarz inequality. Two applications of the Holder inequality yield a bound for $|U{|}^{4k^2}$ containing $Z^{8k^2-4k}{J}_{k,r}(Z{)}^2$ and a product of short geometric-sum bounds over the moment differences. Thus a sharp mean-value estimate combines with rational approximation or spacing of the coefficients to prove cancellation. When all coefficients are integers, $U=\lfloor Z{\rfloor }^2$, so mean-value estimates alone cannot force cancellation.

If $f_n\rightharpoonup f$ in $L^p$, $g_n\to g$ in $L^q$, and $1/p+1/q=1$, thenIndeed, split the difference into the weak pairing with $g$ and a term bounded by the Holder inequality using $g_n-g$.