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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.

Let $V={\mathbb{C}}^n$ and $m\ge 0$. Identify the symmetric power ${\mathrm{Sym}}^{m}V$ with the degree-$m$ part of the symmetric algebra on the basis vectors $e_1,\dots ,e_n$, using formal variables $x_i$ for these vectors. Its monomial basis isThe diagonal maximal torus of the unitary group acts on this monomial by the character $z_1^{a_1}\cdots z_n^{a_n}$. Distinct tuples give distinct characters, so each weight space is one-dimensional.

Suppose $M$ is a nonzero invariant complex subspace. For a nonzero vector $w\in M$, select a monomial with nonzero coefficient and extract it by torus averaging:Fourier orthogonality makes this a nonzero scalar multiple of $x^a$. It belongs to $M$, so $M$ contains a monomial.

Differentiating the unitary group action makes $M$ invariant under $\mathfrak{u}(n)$ and hence under its complexification ${\mathfrak{gl}}_n(\mathbb{C})$. The matrix unit $E_{ij}$ replaces an occurrence of $e_j$ by $e_i$, so its action on the symmetric algebra isWhenever $a_j>0$, this transfers one unit of degree from coordinate $j$ to coordinate $i$ with a nonzero coefficient. Starting with any monomial, such transfers can produce every other tuple of nonnegative integers summing to $m$. Thus $M$ contains every monomial and equals ${\mathrm{Sym}}^{m}V$.

ConsequentlyFor $m=0$ this is the one-dimensional trivial representation, and for $n=1$ every symmetric power is also one-dimensional. This weight-transfer proof of irreducibility of symmetric powers uses neither the Weyl character formula nor the Weyl dimension formula, so no unproved dimension formula enters the argument.

Write the diagonal maximal torus of $U(n)$ as $t=\mathrm{diag}(z_1,\dots ,z_n)$, $|z_i|=1$, and let $\delta =(n-1,n-2,\dots ,0)$. Irreducible complex representations are indexed by dominant integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$, their highest weights. The Weyl character formula in integer-exponent form isThe ratio initially uses distinct $z_i$, and extends to repeated eigenvalues by continuity. Every unitary matrix is conjugate to such a diagonal matrix, so this determines the character on the entire group. The shift $\delta$ differs from the usual Weyl vector $\rho =((n-1)/2,(n-3)/2,\dots ,(1-n)/2)$ by a multiple of $(1,\dots ,1)$; the corresponding determinant factors cancel between numerator and denominator.

We first justify the highest-weight information needed in the proof. Restriction to the torus splits a representation into integer weight spaces. In an irreducible representation choose a lexicographically largest weight $\lambda$ and a nonzero vector $v$ of that weight. A matrix unit $E_{ij}$ with $i<j$ raises the weight by $e_i-e_j$, so it annihilates $v$. The character is invariant under permutations of diagonal coordinates, because permutation matrices conjugate the torus. Hence permutations of $\lambda$ are also weights, and lexical maximality forces ${\lambda }_1\ge \cdots \ge {\lambda }_n$.

Irreducibility implies that $v$ generates the representation under the complexified Lie algebra. Indeed, a Lie-algebra invariant subspace is preserved by exponentials of skew-Hermitian matrices, which generate $U(n)$. Reorder any product of matrix units into lower-triangular units, diagonal units and upper-triangular units, usingInduction on length and on the number of out-of-order pairs gives the required spanning version of the Poincaré-Birkhoff-Witt theorem. Upper-triangular units kill $v$, and diagonal units act by scalars. Every nonempty product of lower-triangular units lowers its weight by a nonzero sum of positive roots. Therefore the top weight space is exactly $\mathbb{C}v$, and its multiplicity is one. This supplies the leading coefficient without assuming a character formula.

For completeness, every dominant integer tuple occurs. Put $d_j={\lambda }_j-{\lambda }_{j+1}\ge 0$ and form the representationIts lexicographically highest weight is $\lambda$, with one-dimensional highest space, represented by the tensor products of $e_1\wedge \cdots \wedge e_j$. A negative determinant exponent is a legitimate one-dimensional unitary-group character. Averaging an inner product over normalized Haar measure gives invariant orthogonal complements, so the tensor representation splits into irreducible summands. One of them has highest weight $\lambda$. Uniqueness will also follow from the formula and Schur orthogonality relations.

Define the alternant $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ for a strictly decreasing integer tuple $\kappa$. The denominatoris the Vandermonde determinant, and its modulus is the denominator modulus in the Weyl integration formula. The Laurent polynomial ${\chi }_{\lambda }{A}_{\delta }$ is alternating. Any alternating Laurent polynomial decomposes as a finite summonomials with repeated exponents have zero coefficient, and the remaining permutation orbits are indexed uniquely by strictly decreasing tuples.

The lexicographically highest monomial in the product is $z^{\lambda +\delta }$ with coefficient one. Indeed, the highest monomial of the character is $z^{\lambda }$ with coefficient one, and the unique lexical maximum among permutations of $\delta$ is $\delta$ itself. It follows that $c_{\lambda +\delta }=1$.

Fourier orthogonality on the torus, whose probability measure is ${\prod }_{i}d{\theta }_i/(2\pi )$, gives alternant orthogonality on a unitary torus:Expand both determinants: an integral survives exactly when the two exponent tuples are identical up to permutation. Strict decreasing order then forces $\kappa =\eta$, and its $n!$ matching terms each contribute one.

Finally, Schur orthogonality relations give ${\int }_{U(n)}|{\chi }_{\lambda }{|}^2=1$. Applying the allowed Weyl integration formula to this class function givesSince the coefficient $c_{\lambda +\delta }$ already equals one, every other coefficient vanishes. Thus ${\chi }_{\lambda }{A}_{\delta }=A_{\lambda +\delta }$, proving the formula. If two irreducible representations had the same highest weight, their characters would now coincide; the Schur orthogonality relations imply that they are isomorphic. Apparent singularities at repeated diagonal eigenvalues are removable because the quotient equals the actual finite Laurent-polynomial character. No dimension formula has been assumed.

We first derive the exterior area theorem for univalent functions directly from Green's theorem, then obtain the coefficient bound needed for the Koebe quarter theorem.

Suppose $G$ is univalent on $|\zeta |>1$ and has a Laurent seriesFor $R>1$, the image of $R{e}^{it}$, $0\le t\le 2\pi$, is a regular simple closed curve. By the Jordan curve theorem the curve has one bounded and one unbounded complementary component. The image of $|\zeta |>R$ lies in the unbounded component, since it is connected, avoids the curve by injectivity and contains points near infinity. It fills that component: any finite boundary limit of the image has preimages bounded by $G(\zeta )/\zeta \to 1$, and a convergent preimage subsequence must land on $|\zeta |=R$ or else give an interior image point. Thus no boundary of the image lies inside that complementary component. The curve is positively oriented about its bounded interior, since the exterior is on the right of $R{e}^{it}$ and a holomorphic map with nonzero derivative preserves orientation. Green's formula gives its enclosed area asFor completeness, the positive Fourier mode is $R{e}^{it}$ and the negative modes are $b_n{R}^{-n}{e}^{-int}$. On multiplying by the conjugate series and integrating, all unequal-frequency products integrate to zero by Fourier orthogonality. The equal-frequency products give $i{R}^2$ and $-in|b_n{|}^2{R}^{-2n}$, respectively; the constant term gives zero. The Laurent series and its differentiated series converge uniformly on the circle, since $R>1$, so these operations are justified. Nonnegativity of $A_R$ yields $\sum n|b_n{|}^2{R}^{-2n}\le R^2$. Letting $R\downarrow 1$, or first doing so for every finite partial sum, givesNo regularity of the map on the unit circle has been assumed.

Now let $f(z)=z+a_2{z}^2+\cdots$ be a normalized univalent function. Its only zero is at zero, so $f(z)/z$, extended there by the value one, is holomorphic and nowhere zero. On the simply connected disc it has a holomorphic square root with value one at zero. Its odd square-root transform isThis is an odd function which is holomorphic and normalized. It is also univalent: $h(z)=h(w)$ gives $f(z^2)=f(w^2)$ and hence $z=\pm w$. If $z=-w$, oddness gives $h(z)=-h(z)$, so $h(z)=0$, forcing $z=w=0$. Thus $G(\zeta )=1/h(1/\zeta )$ is univalent on the exterior disc, with expansionApplying the area bound to its first negative coefficient proves the second coefficient bound for normalized univalent functions, $|a_2|\le 2$.

Let $w_0$ be any point omitted by $f$. It is nonzero. Postcomposition with a Möbius map gives another normalized univalent functionwhose denominator does not vanish. Applying the second coefficient bound both to $f$ and to $f_{w_0}$ givesEvery omitted point therefore has modulus at least $1/4$, provingThis establishes the Koebe quarter theorem from the allowed area formula.

Sharpness is witnessed by the Koebe function $k(z)=z/(1-z{)}^2$. It has the required value and derivative at zero. The equality $k(z)=k(w)$ simplifies to $(z-w)(1-zw)=0$, and $|zw|<1$ makes it injective. Moreover,Thus $-1/4$ is omitted, so no larger universal radius is possible. In fact $4k(z)+1=((1+z)/(1-z){)}^2$ shows its image is the plane slit along $(-\infty ,-1/4]$.

Without univalence take a real $\lambda >4$ andIt is entire, with $f_{\lambda }(0)=0$ and $f_{\lambda }^{\prime }(0)=1$, but it omits $-1/\lambda$, a point of modulus less than $1/4$, since the exponential never vanishes. It is genuinely not univalent on the disc: the distinct points $\pm \pi i/\lambda$ lie there and both give exponential value $-1$. Even its derivative never vanishes, illustrating why local conformality alone does not give the quarter-disc inclusion.

The expectation of a complex random quantity is defined through its real and imaginary parts. Both $\cos X$ and $\sin X$ are bounded, so all expectations below exist. Expand the addition formulae and apply the expectation of a product of independent random variables identity:Combining real and imaginary parts provesFor the simple symmetric random walk, write $W_n={\sum }_{j=1}^n{\xi }_j$, with independent increments satisfying $P({\xi }_j=1)=P({\xi }_j=-1)=1/2$. The usual casino model uses these independent, or conditionally fair, increments. Their characteristic function is $(e^{iu}+e^{-iu})/2=\cos u$, so repeated factorization givesThis is real for real $u$. Equivalently, the distribution of $W_n$ is symmetric, making $E\sin (u{W}_n)=0$.

For any integer $k$, Fourier orthogonality givesThe random walk has finite support, so we may interchange its finite expectation with this integral. ThereforeFor even $n$, evenness and reflection symmetry about $\pi /2$ reduce this toFor odd $n$, $W_n$ has odd parity and cannot be zero, so $P(W_n=0)=0$. As a separate combinatorial check, for $n=2m$ the return probability is $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/2^{2m}$, since exactly $m$ positive increments are required.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

For $d$ variables, integrate over the product of $d$ unit circles:For multi-indices $\alpha ,\beta \in {\mathbb{Z}}_{\ge 0}^d$, applying one-variable Fourier orthogonality in each variable givesIf $f={\sum }_{\alpha }{a}_{\alpha }{z}^{\alpha }$, its squared norm is ${\sum }_{\alpha }|a_{\alpha }{|}^2$, a finite sum. The form is therefore a positive definite Hermitian form andThis is the monomial orthonormal basis on the unit torus.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.

The integral is linear in its first argument and conjugate-linear in its second, and complex conjugation of the integrand shows $⟨g,f⟩=\overline{⟨f,g⟩}$. Thus it is a Hermitian form.

If $f(z)={\sum }_{j=0}^N{a}_j{z}^j$, Fourier orthogonality givesThis is nonnegative and vanishes only when every coefficient vanishes, that is, only for the zero polynomial. Hence the Hermitian form is positive definite.

Use the mixed Dirichlet-Neumann modes on an interval. To justify their Fourier series, extend $y$ evenly across $x=1$ to $[0,2]$, then oddly across $x=0$, and repeat with period four. The condition $y(0)=0$ makes the odd extension continuous, and $y^{\prime }(1)=0$ makes the even reflection continuously differentiable. The resulting periodic function has a convergent Fourier sine series. Reflection about $x=1$ eliminates the even sine harmonics, leavingAlternatively these values follow directly from $\sin {\lambda }_{n}0=0$ and ${\lambda }_n\cos {\lambda }_n=0$. Fourier orthogonality givesThe extension argument also supplies completeness, not merely orthogonality, of this Fourier sine basis.

Expand the forcing as $x\cos \pi x=\sum b_n\sin {\lambda }_{n}x$. Product-to-sum and integration by parts giveEach mode of the differential operator has eigenvalue $-({\lambda }_n^2+{\alpha }^2)$. Hence, for real $\alpha$, the solution isThe coefficients are $O(n^{-4})$, so the series and its first two differentiated series converge uniformly. Therefore it solves the equation and the boundary conditions term by term. Uniqueness follows by multiplying the homogeneous equation by its real solution and integrating: the boundary terms vanish and ${\int }_0^1[(y^{\prime }{)}^2+{\alpha }^2{y}^2],dx=0$, forcing $y=0$.

The Fourier cosine series on $[0,L]$ isIt is the Fourier series of the even extension to $[-L,L]$, repeated with period $2L$. The Fourier sine series isIt comes from the odd extension and its $2L$-periodic extension. These Fourier coefficient formulas follow by Fourier orthogonality, since every nonconstant sine or cosine has squared integral $L/2$ on $[0,L]$, while the constant function has squared integral $L$.

For $\alpha =3$, let $F_n$ denote the time factor normalized by $F_n(0)=0$, $F_n^{\prime }(0)=1$. By superposition the Fourier sine series solution isAt $t=0$, its initial velocity is ${\sum }_{n\ge 1}{b}_n\sin (nx)={\sin }^{2}x$. Multiply by $\sin (mx)$, integrate on $[0,\pi ]$, and use the permitted interchange and Fourier orthogonality. This yields the requested integral expressionUnder $x\mapsto \pi -x$, ${\sin }^{2}x$ is unchanged and $\sin (n(\pi -x))=(-1{)}^{n+1}\sin (nx)$. Hence every even-index Fourier coefficient is zero, including $b_2$.

For odd $n$, use ${\sin }^{2}x=(1-\cos 2x)/2$ and the product-to-sum formula:ThusIn particular $b_1=8/(3\pi )$ and $b_3=-8/(15\pi )$. An explicit version of the solution of a differential equation isOnly the first, overdamped normal mode, the third, critically damped normal mode, and the odd underdamped normal modes of index at least five occur. No odd-index Fourier coefficient vanishes.

For a nonzero integer $k$, direct integration giveswhereas ${\int }_{-\pi }^{\pi }1dx=2\pi$. The product-to-sum formulastherefore establish Fourier orthogonality: for positive integers $m,n$,Here ${\delta }_{mn}$ is the Kronecker delta. In particular the constant function is orthogonal to every nonconstant sine and cosine in this system. Integrating the given Fourier series, then multiplying it by $\cos (mx)$ or $\sin (mx)$ and integrating, isolates its Fourier coefficients. Using the permitted interchange of the sum and integral givesIf $f$ is an even function, then $f(x)\sin (nx)$ is an odd function, while $f(x)\cos (nx)$ is an even function. ThusThe Fourier series consequently reduces to a Fourier cosine series.

The Fourier cosine series and Fourier sine series use the respective formsTheir Fourier coefficients, obtained from Fourier orthogonality, areIn particular the constant term is $a_0/2={\pi }^{-1}{\int }_0^{\pi }f(t)dt$. The Fourier cosine series is the Fourier series of the even extension $f(|x|)$, made $2\pi$-periodic. The Fourier sine series is the Fourier series of the odd extension, equal to $f(x)$ for $0<x<\pi$ and $-f(-x)$ for $-\pi <x<0$, again made $2\pi$-periodic.

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

Take $\boxed{f(x)=\sin x}$. This function is not a polynomial: it is nonzero and satisfies $f^{\prime \prime }=-f$, whereas a nonzero polynomial cannot satisfy that identity, since taking two derivatives reduces its degree or gives zero.

By Fourier orthogonality, the Fourier sine series consists of the single term $\sin x$. Thus every truncation with $N\ge 1$ is exact, including at the endpoints. In contrast, using the product-to-sum trigonometric identity in its Fourier cosine coefficients givesFor example, ${\int }_0^{\pi }\sin x\cos (2mx)dx=-2/(4m^2-1)$ follows by integrating the two resulting sine terms. Therefore the Fourier cosine series isEvery finite cosine truncation omits infinitely many nonzero Fourier coefficients, so Parseval's identity gives it strictly positive squared L2 norm error. The one-term sine representation has zero error in both the L2 norm and the supremum norm, proving the requested advantage.

Set $u=X(x)Y(y)$ in Laplace's equation. Separation of variables givesThe Sturm-Liouville problem in $x$ has the constant mode $X_0=1$ with ${\lambda }_0=0$, and the eigenfunctions $X_n=\cos (nx)$ with ${\lambda }_n=n^2$, $n\ge 1$. Indeed, integration by parts makes $\lambda$ nonnegative; for $\lambda >0$ the first Neumann boundary condition removes the sine term and the second requires $\sqrt{\lambda }$ to be an integer. For $\lambda =0$, both derivative conditions make $X$ constant.