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A bounded function belongs to $C^{0,\alpha }({\mathbb{R}}^n)$ when

Take ${\Omega }^{\prime }$ nonempty and compactly contained in $\Omega$. The argument is a compactness proof of a nonnegative elliptic solution estimate, with the global Schauder estimate retaining control all the way to the boundary. The operator and the Hölder space exponent are fixed throughout.

Suppose the estimate fails. For each integer $k$ there is an admissible nonnegative solution $u_k$, with forcing $f_k$, such thatPut $v_k=u_k/M_k$ and $g_k=f_k/M_k$. Then $0\le v_k\le 1$, ${\max }_{\overline{\Omega }}{v}_k=1$, $v_k=0$ on the boundary, $L{v}_k=g_k$, andGlobal elliptic regularity and the global Schauder estimate for zero Dirichlet boundary condition giveThis estimate includes the $C^0$ term, so it does not require invertibility of $L$ or a sign restriction on $c$. The compact embedding of Hölder spaces, or the Arzela-Ascoli theorem applied through second partial derivatives, gives a subsequence converging in $C^2(\overline{\Omega })$ to $v$. Thus $v\ge 0$, $Lv=0$, $v=0$ on the boundary, and $\max v=1$.

Choose $x_k\in {\Omega }^{\prime }$ with $v_k(x_k)\le {\inf }_{{\Omega }^{\prime }}{v}_k+1/k$. A further subsequence has $x_k\to x_{\ast }\in \overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence yields $v(x_{\ast })=0$. Part (b) says that a nonnegative homogeneous solution on the connected domain is either zero everywhere or strictly positive inside. Both alternatives contradict, respectively, $\max v=1$ and $v(x_{\ast })=0$. ThereforeThe interior infimum and the forcing control the global supremum. Global boundary control is crucial here: an interior-only compactness argument could lose the normalized maximum at the boundary.

Use the supremum norm barrier for an elliptic Dirichlet problem to obtain a domain-only constant $C_{\Omega }=e^{2d}$. Split the coefficient as $c=c_{-}+c_{+}$, where $c_{-}\le 0$ and $c_{+}=\max (c,0)$. Both parts are Hölder continuous functions. ChooseIf $u$ solves the homogeneous zero-boundary problem for $\Delta +c$, thenPart (a), applied to this nonpositive zeroth-order coefficient, givesHence $u=0$. The Fredholm alternative for an elliptic Dirichlet problem now gives existence and uniqueness for every forcing and boundary datum in the stated Hölder spaces. This proves the small positive zeroth-order perturbation of a Dirichlet problem without imposing a bound on the negative part of $c$. The value of $ϵ$ need not be optimal.

For $0<\alpha \le 1$, uniform Lipschitz-$\alpha$ means that one constant $M$ satisfies $|f(x)-f(y)|\le M|x-y{|}^{\alpha }$ for every $x,y\in [0,1]$. The resulting Hölder space has normFor the subsequent estimates at general $\alpha >0$, use the higher-order Hölder class convention: write $\alpha =r+\beta$ with $r\ge 0$ an integer and $0<\beta \le 1$, and require $f\in C^r([0,1])$ with $f^{(r)}$ uniformly Lipschitz-$\beta$. ThusAt integer $\alpha =k$, this uses $C^{k-1,1}$; the alternative classical $C^k$ convention is stronger on this compact closed interval and also suffices for the estimates.

For exponents above one, a first-difference inequality alone forces the function to be constant. Partition $[x,y]$ into $n$ equal pieces and use the triangle inequality: $|f(y)-f(x)|\le M|y-x{|}^{\alpha }{n}^{1-\alpha }\to 0$. Consequently a nontrivial higher-order definition is needed for the full range of exponents in the later wavelet estimates. Equivalently, higher-order Hölder class gives a local Taylor polynomial with remainder bounded by $C\Vert f{\Vert }_{C^{\alpha }}|x-x_0{|}^{\alpha }$.

Take ${\Omega }^{\prime }$ nonempty. If the bound failed, normalization would give nonnegative $v_j$ with supremum one, zero boundary values, ${\inf }_{{\Omega }^{\prime }}{v}_j\to 0$, and $\Vert L{v}_j{\Vert }_{C^{0,\alpha }}\to 0$. The global Schauder estimate bounds them in the Hölder space $C^{2,\alpha }$. By compact embedding of Hölder spaces, a subsequence converges in $C^{2,\beta }$, $0<\beta <\alpha$, to $v\ge 0$ with $Lv=0$, zero boundary data and supremum one.

Points approaching the infima have a subsequence converging in $\overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence gives an interior zero of $v$, and part (c), the strong minimum principle for elliptic operators after replacing $c$ by $\min (c,0)$, gives $v=0$. This contradicts its supremum one, proving

Normalize $\hat{\phi }(0)=1$, as in the displayed product convention. Iterating the scaling refinement equation gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$. The finite-product identityhas the value one at zero by continuity. ThereforeWrite $M=\sup |L|$. Since $L(0)=1$, we have $M\ge 1$. For $|\xi |>1$, take $J=\lceil {\log }_2|\xi |\rceil$. The first $J$ factors are bounded by $M^J\le M|\xi {|}^{{\log }_{2}M}$. For the remaining factors, the trigonometric polynomial satisfies $|L(t)|\le 1+C|t|$ on $|t|\le 1$, henceUsing $|(1-e^{-i\xi })/(i\xi )|\le \min (1,2/|\xi |)$ givesThe strict hypothesis permits $\epsilon >0$ with $N-{\log }_{2}M\ge \alpha +1+\epsilon$. The given Fourier-decay criterion proves uniform Hölder regularity of exponent $\alpha$. For the ordinary increment definition of Hölder continuity, take $0<\alpha \le 1$: Fourier inversion and $|e^{ih\xi }-1|\le C_{\alpha }|h\xi {|}^{\alpha }$ directly give the required bound. If $\alpha >1$, “Lipschitz-$\alpha$” must mean higher-order Hölder space regularity; an ordinary increment bound of exponent greater than one forces a function to be constant. An unnormalized scaling function contributes its constant phase to the product formula.

Taylor's theorem and $g(0)=g^{\prime }(0)=0$ giveCombining these with the product estimate in a Hölder space yieldsWritingand using the same product estimate gives

Fix $B_R(x_0)\subset B_1$, put $A_0=A(x_0)$, and let $v$ have the same boundary values as $u$ while solving $\mathrm{div}(A_0\nabla v)=0$. Then $w=u-v\in W_0^{1,2}(B_R)$ satisfiesSince $u\in C^1$ gives $|Du|\le L$ locally, the supplied energy estimate and Hölder continuity of $A$ yieldThe constant-coefficient decay estimate givesThus, for $\Phi (r)={\int }_{B_r}|\nabla u-(\nabla u{)}_{B_r}{|}^2$,Because $n+2>n+2\alpha$, the Campanato iteration lemma gives $\Phi (\rho )\le C{\rho }^{n+2\alpha }$ on balls in $B_{1/2}$. Hence $\nabla u\in {\mathcal{L}}^{2,n+2\alpha }=C^{0,\alpha }$ and $u\in C^{1,\alpha }(B_{1/2})$.

A function with finite Hölder seminorm of exponent $\alpha >1/2$ has squared-increment sum on a uniform $N$-interval partition bounded by $C^2{N}^{1-2\alpha }$, hence tending to zero. This conflicts with a nonzero quadratic-variation limit. In particular dyadic quadratic variation of Brownian motion prevents a Brownian coordinate law supported on such a Hölder space. This argument alone gives no conclusion at the critical exponent $1/2$.