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For $g(x)={\sum }_{k\ge 0}{5}^{-k}\cos (5^{k}x)$, all frequencies satisfy $5^k\equiv 1(\mathrm{mod}4)$. At $x=\pi /2$, the increment is $-{\sum }_{k\ge 0}{5}^{-k}\sin (5^{k}h)$. If $5^m\le 1/h<5^{m+1}$, the first $m+1$ terms have $0<5^{k}h\le 1$ and contribute at least $(m+1)h\sin 1$, while the remaining tail has magnitude at most $5^{-m}/4\le 5h/4$. This proves a lower modulus of continuity bound of order $h\log (1/h)$ for small $h$. The inverse theorem for trigonometric approximation supplies the matching upper bound because the positive lacunary trigonometric series has $E_n(g)\asymp n^{-1}$.

The first Jackson theorem for periodic approximation applied to $|\sin \theta |$ gives algebraic best error $O(n^{-1})$ for $\sqrt{1-x^2}$. Its ordinary interval modulus of continuity at $1/n$ is at least $\sqrt{2/n-1/n^2}$. Substituting that error rate into the unmodified inverse theorem for trigonometric approximation would instead bound the modulus by $O(\log (n+1)/n)$, a contradiction. Algebraic inverse estimates must incorporate the endpoint compression of cosine substitution for polynomial approximation.

The normalized Fejér kernel satisfies $0\le F_n(t)\le \min \{n/(2\pi ),\pi /(2n{t}^2)\}$ on $0<|t|\le \pi$. Splitting its first absolute moment at $1/n$ gives $\int |t|F_n(t)dt\le 1/(2\pi n)+(\pi /n)\log (\pi n)$. The chaining inequality for the modulus of continuity, $\omega (f,t)\le (1+t/\delta )\omega (f,\delta )$, then bounds the Fejér sum error by $\omega (f,\delta )[1+{\delta }^{-1}\int |t|F_n]$. Taking $\delta =\log n/n$ proves the stated bound for $n\ge 2$ and hence uniform convergence for continuous periodic functions. The logarithm vanishes at $n=1$, so the displayed estimate is not intended at that index.

Every continuous $2\pi$-periodic function admits degree-at-most-$n$ trigonometric polynomials with supremum norm error bounded by a universal constant times its modulus of continuity at scale $1/n$, for $n\ge 1$. This first-order Jackson-type estimate transfers to algebraic polynomials through cosine substitution for polynomial approximation.

At each point there is at most one supported Haar wavelet per scale. If a detail coefficient is below $\delta$, its value there is at most $\delta 2^{j/2}/\sqrt{L}$. Summing these bounds up to level $K$ gives the displayed geometric estimate. Selecting $K$ so the final term is of the order of the modulus of continuity controls all omitted fine details, even though arbitrary deletions from a uniformly convergent series need not preserve convergence.

A Jackson-type estimate bounds an approximation error by a smoothness quantity such as a modulus of continuity, with the smoothness evaluated at a scale comparable to the reciprocal of the approximating degree.

Between two prescribed values, linear interpolation assigns $(1-t)f(a)+tf(b)$ at $(1-t)a+tb$, for $0\le t\le 1$. Joining successive samples produces a piecewise linear function. For a continuous function on a compact interval, interpolants along grids with mesh tending to zero converge in the uniform norm; their error is bounded by the modulus of continuity at the mesh size.

For nonnegative partition-normalized order-$k$ B-splines, let $Vf(t)={\sum }_{i}f({\tau }_i)N_i(t)$ with each sample point inside $[t_i,t_{i+k}]$. If $N_i(t)$ is nonzero, both $t$ and ${\tau }_i$ lie in an interval of length at most $k|\Delta |$. Their function values differ by at most the modulus of continuity at that length. Summing with nonnegative weights of total one proves the displayed bound. With fixed order and vanishing maximum knot gap, uniform continuity gives uniform convergence. The same argument works for bounded piecewise-continuous outputs if full interior knot multiplicities are allowed; their individual continuity should not then be asserted.

Set $\mu (0)=0$ and extend $\mu (r)=1$ for $r\ge 1$. This positive, nondecreasing modulus of continuity is weaker than a linear Lipschitz continuity bound but still has ${\int }_0^{1}dr/\mu (r)=\infty$. At small distances it is equivalent to $r\log (1/r)$. A signed $r\log r$ cannot be an upper modulus at distances less than one. The planar vorticity velocity kernel maps $L^1\cap L^{\infty }$ into velocity fields with this modulus.

The original PDF's kernel has an incorrect normalization. With ordinary $dt$, its integral is $\pi$, not one. Already at $n=1$ the printed kernel is the constant $1/2$. For $f\equiv 1$, the printed operator therefore returns $\pi$, while $\omega (f,\delta )=0$ for every $\delta$. Thus the literal inequality with the printed kernel is false. The Fejér kernel representing the stated average of Fourier partial sums isThe ratio at zero is interpreted by its limit. To verify the normalization and the averaging property, use the finite geometric sum:Fourier orthogonality gives unit integral; the displayed Fourier coefficients are exactly the multipliers of the Fejér sum. This also proves nonnegativity.

For the correctly normalized Fejér sum, subtraction of $f(x)$ givesTake $0<\delta \le 1$ and split at $|t|=\delta$. The inner part is at most $\omega (f,\delta )$, because the kernel is nonnegative and has mass one. The strict inequality in the definition of the modulus of continuity causes no endpoint issue: continuity gives the same bound at distance exactly $\delta$.

On $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$, soDivide an arc of length $|t|$ into at most $1+|t|/\delta$ shorter arcs. The triangle inequality then bounds $|f(x-t)-f(x)|$ by $(1+\pi /\delta )\omega (f,\delta )$. ConsequentlyChoosing $\delta =n^{-1/2}$ proves the requested fractional-scale Fejér approximation bound for the intended normalization, for example with the absolute constant $1+\pi +{\pi }^2$:The same proof applies to complex-valued continuous functions, since the estimates use absolute values.

The original PDF prints $O(n^{\alpha })$, with a positive exponent. That premise is insufficient; the intended decay rate is $O(n^{-\alpha })$. Indeed every bounded continuous periodic function has $E_n(f)\le \Vert f{\Vert }_{\infty }=O(n^{\alpha })$. For an explicit counterexample, on $[-\pi ,\pi ]$ setand extend periodically. This is continuous, but $\omega (h,\delta )\ge 1/\log (e/\delta )$ for $0<\delta <1$, which is not $O({\delta }^{\alpha })$ for any $\alpha >0$. Thus the sign correction is required in the actual printed question, not just in its converted TeX.

For real continuous $2\pi$-periodic $f$, write $E_n$ for best uniform approximation by degree-at-most-$n$ trigonometric polynomials and $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. A quantitative inverse theorem for trigonometric approximation iswhere $C$ is universal. Here is a direct reason for the sum. Choose best approximants $p_j$ and $2^r\le n<2^{r+1}$. WriteThe Bernstein inequality for trigonometric polynomials givesFor $j\ge 2$, monotonicity of best errors bounds $2^{j+1}{E}_{2^{j-1}}$ by $8{\sum }_{\nu =2^{j-2}}^{2^{j-1}-1}{E}_{\nu }$. The first terms are bounded by a fixed multiple of $E_0$. Also $2E_{2^r}\le 4n^{-1}{\sum }_{\nu =0}^n{E}_{\nu }$. Nowproves the theorem, for example with a sufficiently large absolute constant $C=20$.

Suppose, as intended, $E_{\nu }(f)\le A{\nu }^{-\alpha }$ for $\nu \ge 1$, after enlarging $A$ to include any finite exceptional indices, where $0<\alpha <1$. Since ${\sum }_{\nu =1}^n{\nu }^{-\alpha }\le n^{1-\alpha }/(1-\alpha )$, the inverse theorem givesFor $0<\delta \le 1$ take $n=\lfloor 1/\delta \rfloor$. Then $1/(n+1)<\delta \le 1/n$, so monotonicity of the modulus of continuity and $n^{-\alpha }\le 2^{\alpha }{\delta }^{\alpha }$ give $\omega (f,\delta )\le 2^{\alpha }{C}_f{\delta }^{\alpha }$. For $\delta >1$, the separate bound $\omega (f,\delta )\le 2\Vert f{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }{\delta }^{\alpha }$ covers the remaining scales. Therefore the corrected conclusion isNow put $\beta =\log 2/\log 3$. The given best approximant to the Weierstrass function leaves the uniformly convergent tail ${\sum }_{k=m+1}^{\infty }{2}^{-k}\cos (3^{k}x)$. Its supremum norm is exactly $2^{-m}$, attained at $x=0$; henceThe corrected inverse theorem gives $\omega (g,\delta )=O({\delta }^{\beta })$. To show that this order is sharp, set $h_m=\pi /3^{m+1}$. Every term of $g(0)-g(h_m)$ is nonnegative, and for $k\ge m+1$ the cosine equals $-1$ because $3^{k-m-1}$ is odd. ThusGiven small $\delta >0$, select $m$ with $h_m\le \delta <3h_m$. Then $\omega (g,\delta )\ge 4(3\pi {)}^{-\beta }{\delta }^{\beta }$. Combining the estimates gives the precise small-scale orderThe exact approximation error can also be verified independently of the supplied fact: at $x_j=j\pi /3^{m+1}$ the tail has alternating values $(-1{)}^j{2}^{-m}$. There are enough such points in one period for the trigonometric Chebyshev alternation theorem, since $n<3^{m+1}$. This is the positive lacunary trigonometric series mechanism behind the best partial sums.

If polynomials with integer monomial coefficients converge uniformly to $f$, their values at zero and one are integers converging to $f(0)$ and $f(1)$. A convergent integer sequence is eventually constant: once all its terms lie in an interval of length less than one they must agree. Hence both endpoint values are integers.

Conversely assume those values are integers. The rounded Bernstein polynomial is an integer-coefficient polynomial, because its displayed scalar coefficients are integers and each $x^k(1-x{)}^{n-k}$ expands with integer coefficients. Part b gives its vanishing distance from $B_{n}f$. For completeness, the remaining uniform convergence $B_{n}f\to f$ follows directly from uniform continuity. Let $X$ have the binomial distribution with parameters $n,x$; then $B_n(f,x)=\mathbb{E}f(X/n)$, $\mathbb{E}(X/n)=x$, and $\mathrm{Var}(X/n)=x(1-x)/n\le 1/(4n)$. Splitting into $|X/n-x|\le \delta$ and its complement, and applying Chebyshev's inequality, giveswhere $\omega$ is the modulus of continuity. First choose $\delta$ to make the first term small uniformly, then take $n$ large. The triangle inequality now gives $\Vert B_n^{\ast }f-f{\Vert }_{\infty }\to 0$. We have proved the integer-coefficient polynomial approximation characterization

Use the periodic modulus of continuitySubdividing a displacement $t$ into at most $\lceil |t|/\delta \rceil$ steps of size at most $\delta$ proves the chaining inequalityFor $|t|\le \delta$, the sharper bound $\omega (f,\delta )$ holds directly.

Set $\delta =n^{-\alpha }$, where $0<\alpha <1$ and $n\ge 1$. Positivity and unit normalized mass of the Fejér kernel giveThe portion $|t|\le \delta$ is at most $\omega (f,\delta )$. For $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$ and $|\sin (nt/2)|\le 1$, soApply this bound and the chaining inequality to the two tails:Let $\beta =1-\alpha >0$. Calculus gives ${\sup }_{u\ge 1}{u}^{-\beta }\log u=1/(e\beta )$, attained at $u=e^{1/\beta }$, and $u^{-\beta }\le 1$. Thus a possible explicit constant for the fractional-scale Fejér approximation bound isThe constant depends only on $\alpha$, not on $f$ or $n$. Since a continuous periodic function is uniformly continuous, the right-hand side tends to zero. The logarithmic tail estimate is what permits every $\alpha <1$; a crude global-oscillation bound would not prove the whole requested range.

Put $L=2\pi$ and represent the circle by $[0,L)$. The full orthonormal basis of Haar wavelets includes the constant $H_{\ast }=L^{-1/2}$ and, for every dyadic interval $I$ of length $\ell =L{2}^{-j}$, the normalized detailTake half-open intervals and the corresponding values at their boundaries, so every point belongs to exactly one interval at each level. The constant basis function is necessary: if it were omitted, even $f\equiv 1$ could not satisfy the asserted convergence. We use the full conventional Haar wavelet system in the threshold sum.

Let $A_{J}f$ be the function equal to the mean of $f$ on each level-$J$ dyadic interval. Comparing the means on a parent and its two children proves the Haar refinement identity directly: the difference between the child mean and the parent mean is the appropriate value of $c_I{H}_I$. Inductively this gives the Haar projectionWriting ${\omega }_f(s)={\sup }_{\mathrm{dist}(x,y)\le s}|f(x)-f(y)|$ for the modulus of continuity, the cell-average formula yieldsFor a detail on a cell $I$, its integral is zero. Subtracting $f(x_I)$ for any $x_I\in I$ therefore gives the coefficient estimateIn particular $|c_I|\le 2\Vert f{\Vert }_{\infty }\sqrt{\ell }$, so for every $\delta >0$ only finitely many details can have $|c_I|\ge \delta$. The hard-threshold sumis thus a well-defined finite sum. Its chosen details need not form a complete set of levels; this is why convergence of the Haar projections alone does not prove the result.

Fix $\eta >0$. Choose $J\ge 1$ so that ${\omega }_f(L{2}^{-J})\le \eta$. All finer details satisfy $|c_I|\le \eta \sqrt{L}{2}^{-j/2}$ at level $j\ge J$. For sufficiently small $\delta$, every nonzero coefficient at levels below $J$, and the constant coefficient if nonzero, is retained. Also choose the unique integer $K\ge J$ such thatFor $j>K$ we have $|c_I|\le \eta \sqrt{L}{2}^{-j/2}<\delta$, so no such detail is retained. Consequently $T_{\delta }f$ differs from $A_{K+1}f$ only by omitted details at levels $J$ through $K$. At any point there is at most one supported detail per level, and an omitted detail has magnitude at most $\delta /\sqrt{L{2}^{-j}}$. The geometric bound for omitted Haar details givesSince $\Vert A_{K+1}f-f{\Vert }_{\infty }\le \eta$, we concludeThe right side can be made arbitrarily small, proving uniform hard-threshold convergence of normalized Haar expansions:The square-root scaling imposed by the specified $L^2$ normalization is what makes the omitted contributions a controlled geometric series.

For a function on an interval, the modulus of continuity is $\omega (f,t)={\sup }_{|x-y|\le t}|f(x)-f(y)|$, with the pair restricted to that interval. For a periodic function the equivalent definition is ${\sup }_{\theta ,|h|\le t}|g(\theta +h)-g(\theta )|$. The mean value theorem and $|\sin \theta |\le 1$ giveBoth cosine values lie in $[-1,1]$. Hence for $|h|\le t$,Taking the supremum provesIf the periodic modulus of continuity is defined using circular distance instead, choose lifts at that shortest distance and apply the same argument. The cosine substitution for polynomial approximation therefore does not enlarge this modulus.

Use the modulus of continuity on the circle, $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. Splitting a displacement into steps of length at most $\delta$ and applying the triangle inequality givesThe Fejér kernel is nonnegative and has integral one, soWe therefore need its first absolute moment. The inequalities $|\sin (nt/2)|\le n|\sin (t/2)|$ and $\sin (|t|/2)\ge |t|/\pi$ on $[-\pi ,\pi ]$ implySplitting at $1/n$ givesTaking $\delta ={\delta }_n=(\ln n)/n$ in the preceding error bound proves the Fejér first-moment approximation bound:The constants do not depend on $f$ or $n$. By the Heine-Cantor theorem, a continuous function on the compact circle is uniformly continuous, hence $\omega (f,{\delta }_n)\to 0$ as ${\delta }_n\to 0$. Thus the Fejér sums converge uniformly to $f$. The index restriction is necessary: at $n=1$ the printed scale is zero, while ${\sigma }_{1}f$ is the mean of $f$ and generally differs from it.

The Weierstrass M-test gives a continuous periodic Weierstrass function, since $\sum a^{-k}<\infty$. For $5^m\le n<5^{m+1}$, the given best approximant leaves the tailIts supremum norm is at most the sum of its coefficients, and at $x=0$ all cosine terms equal one, attaining that bound. ConsequentlyPut $\alpha =\ln a/\ln 5$. Then $a^{-m}=(5^m{)}^{-\alpha }$ and $n/5^m\in [1,5)$, giving $E_n(g)\asymp n^{-\alpha }$, with constants depending only on $a$.

For $1<a<5$, part (a) gives $\omega (g,1/n)\le C_a{n}^{-\alpha }$. The first Jackson theorem for periodic approximation, $E_n(g)\le C\omega (g,1/n)$, supplies the matching lower bound. Hence the modulus of continuity has the precise orderAt $a=5$, the inverse theorem for trigonometric approximation gives the upper bound $C\delta \ln (1/\delta )$, and part (c) proves the matching lower bound. Thus the critical Weierstrass modulus isTo pass from scales $1/n$ to all small $\delta$, use monotonicity of the modulus of continuity and adjacent reciprocals, which are comparable to $\delta$.

Take $a=5$ and $x=\pi /2$. Since $5^k\equiv 1(\mathrm{mod}4)$, every cosine term vanishes at $x$, andSet $h=1/n$ and $m=\lfloor {\log }_{5}n\rfloor$. For $0\le k\le m$, the sine argument lies in $[0,1]$, where $\sin u\ge (\sin 1)u$. The initial part therefore has a common sign and magnitude at least $(m+1)(\sin 1)/n$. The remaining tail has absolute value at mostThe reverse triangle inequality now givesfor sufficiently large $n$, because $m+1>\ln n/\ln 5$. This establishes the sharp logarithmic lower boundThe constant can be reduced to include the finitely many small indices: for $n\ge 5$ the displayed lower bound is positive; for $n=2,3$ the first sine exceeds the tail bound $1/4$, and for $n=4$ the first two sines contribute $\sin (1/4)+\sin (5/4)/5$, exceeding the remaining tail bound $1/20$. At $n=1$ the logarithmic right side is zero.

Here $E_n(g)\asymp 1/n$ but $g$ is not Lipschitz continuous, so a first modulus of continuity bound $O(1/n)$ is sufficient by the first Jackson theorem for periodic approximation but is not necessary. Nor is the weaker necessary bound $O(\ln n/n)$ sufficient: an explicit comparison is the logarithmic cusp with slow trigonometric approximationextended periodically. On $[0,\pi ]$ it is increasing and concave, so its largest increment over distance $\delta$ is from zero, giving $\omega (h,\delta )=\delta \ln (e\pi /\delta )$ for $0<\delta \le \pi$.

For a concrete approximation lower bound, its cosine Fourier coefficients satisfyIndeed, integration by parts once replaces $h$ by $h^{\prime }(x)=\ln (\pi /x)$; integrating that against sine with primitive $(1-\cos jx)/j$ yields the stated positive integral. Its logarithmic growth follows because ${\int }_1^A\cos u/udu$ stays bounded by the Dirichlet test. All $a_j$ are negative and absolutely summable.

Let $V_n=2{\sigma }_{2n}-{\sigma }_n$, a de la Vallée Poussin sum. Its multiplier is one for $j\le n$, $2-j/n$ for $n<j<2n$, and zero for $j\ge 2n$; the Fejér summation is a uniform-norm contraction property gives $\Vert V_n\Vert \le 3$. Thus the functional $L_n(f)=f(0)-V_{n}f(0)$ has norm at most four and vanishes on every degree-at-most-$n$ trigonometric polynomial. Because the remaining multiplier weights are nonnegative and all $a_j<0$,The first Jackson theorem for periodic approximation gives the corresponding upper bound. The same first-modulus order $\delta \ln (1/\delta )$ occurs with both $E_n\asymp 1/n$ and $E_n\asymp \ln n/n$. Therefore first-modulus rates alone do not characterize the endpoint approximation class, unlike the matched direct and inverse rates for $0<\alpha <1$.