The weak convergence of distributions $u_j\to u$ meansFor $\theta \in \mathcal{D}(\mathbb{R})$, its convolution with a distribution is the smooth functionThe subscript records the variable on which the distribution acts. With $\check{\theta }(x)=\theta (-x)$, its regular distribution satisfies the useful pairing identityIndeed integrating $\theta (x-y)\varphi (x)$ in $x$ gives $(\check{\theta }\ast \varphi )(y)$; continuity of the distribution justifies moving the pairing through that integral, by approximation with sums of test functions in one common compact support.

Part (i), applied to the reflected mollifier, gives ${\check{\rho }}_{\epsilon }\ast \varphi \to \varphi$ in the space of test functions. Continuity of $u$ therefore provesReflection is harmless even if the chosen standard mollifier is not even.

For the mollifiers themselves, a change of variable givesThus $\boxed{{\rho }_{\epsilon }\to {\delta }_0\text{ in }{\mathcal{D}}^{\prime }(\mathbb{R})}$, with limit the Dirac delta distribution, despite the failure of convergence in the L1 space.

Convergence ${\varphi }_j\to \varphi$ in the space of test functions means that all supports lie in one compact set $K\subset \mathbb{R}$ and $\Vert {\varphi }_j^{(r)}-{\varphi }^{(r)}{\Vert }_{\infty }\to 0$ for every integer $r\ge 0$. An eventual common compact support is equivalent, since finitely many initial supports can be added to $K$.

Write the standard mollifier as ${\rho }_{\epsilon }(x)={\epsilon }^{-1}\rho (x/\epsilon )$, where $\rho \in \mathcal{D}(\mathbb{R})$ is nonnegative, supported in $[-1,1]$, and has integral one. For $0<\epsilon \le 1$, the convolution ${\rho }_{\epsilon }\ast \varphi$ has support in the fixed compact set $\mathrm{supp}\varphi +[-1,1]$. The given convolution derivative identity yieldsEach ${\varphi }^{(r)}$ is uniformly continuous; more explicitly, the mean value theorem bounds the absolute value by $\epsilon \Vert {\varphi }^{(r+1)}{\Vert }_{\infty }\int |y|\rho (y)dy$, uniformly in $x$. Thus

If ${\varphi }_j\to 0$ in the space of test functions and $K$ is a common compact support, then $\Vert {\varphi }_j{\Vert }_{L^1}\le |K|\Vert {\varphi }_j{\Vert }_{\infty }\to 0$, where $|K|$ is its Lebesgue measure.

The mollifiers themselves have no limit in the L1 space. If ${\rho }_{\epsilon }\to g$ in that norm, then for each $\delta >0$ the integral of $|g|$ on $\{|x|>\delta \}$ is zero, because ${\rho }_{\epsilon }$ vanishes there for $\epsilon <\delta$. Hence $g=0$ almost everywhere. This contradicts $\Vert {\rho }_{\epsilon }{\Vert }_{L^1}=1$. Therefore $\boxed{{\rho }_{\epsilon }\text{ does not converge in }{L}^1(\mathbb{R})}$.

No. Consider the continuous even functionThe formulas agree at $|x|=e^{-1}$, and $f(x)\to 0$ at zero. Part (i) defines its even-weight principal-value distribution. However,If this were a regular distribution represented by $g\in L_{\mathrm{loc}}^1(\mathbb{R})$, testing on either side of zero would force $g=f/x$ almost everywhere there, by uniqueness of the locally integrable function representing a regular distribution. Changing the value at zero cannot remove the displayed divergence, contradicting local integrability of $g$.

More generally, the exact condition isNecessity is the preceding argument; sufficiency follows because $f/x$ is then a locally integrable function and its truncated integrals converge absolutely. The value $f(0)=0$ alone provides no such rate of decay.

Choose $R>0$ with $\mathrm{supp}\varphi \subset [-R,R]$. Evenness of $f$ changes the symmetrically truncated integral intoThe mean value theorem gives $|\varphi (x)-\varphi (-x)|\le 2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Since $f$ is continuous, it is bounded on $[0,R]$. Thus the Cauchy principal value exists andThis expression is linear in the test function and satisfiesFor each fixed compact support, choose one $R$ once and for all. The estimate proves that the even-weight principal-value distribution has order of a distribution at most one.

If $\varphi (0)=0$, the mean value theorem also gives $|\varphi (x)|\le |x|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Hence $f(x)\varphi (x)/x$ is absolutely integrable near zero and has compact support. The truncated integrals tend to its ordinary Lebesgue integral, so

The bound $0\le d_n\le 2^{-n-1}$ proves uniform convergence of the series by the Weierstrass M-test. Each summand is continuous and satisfies $d_n(x+1)=d_n(x)$, so its sum $t$, the Takagi function, is continuous and has period one.

For $n\ge m$, both $2^n{u}_m$ and $2^n{v}_m$ are integers, giving $d_n(u_m)=d_n(v_m)=0$. ThereforeThe corners of $d_n$ occur on the grid $2^{-n-1}\mathbb{Z}$. When $n<m$, that grid is contained in $2^{-m}\mathbb{Z}$, so the interior of $[u_m,v_m]$ contains no corner of $d_n$. Its slope on this interval is consequently $+1$ or $-1$.

The interval $[u_{m+1},v_{m+1}]$ is a half of $[u_m,v_m]$. All the old slopes persist on this half, and the new summand contributes one more slope in $\{\pm 1\}$. ThusIf $t$ had a finite derivative $L$ at $x_0$, write $t(y)=t(x_0)+L(y-x_0)+r(y)$ with $r(y)=o(|y-x_0|)$. Since $u_m\le x_0<v_m$ and both endpoints tend to $x_0$,Indeed $|v_m-x_0|+|u_m-x_0|=v_m-u_m$; when $u_m=x_0$, its remainder is zero. This is the secants straddling a differentiability point argument, including dyadic $x_0$. But the displayed increments prevent $S_m$ from converging. Hence $\boxed{t\text{ is nowhere differentiable}}$, using periodicity to cover all real $x_0$.

For the distributional derivative, set $r_n(x)=s(2^{n}x)$, using $s$ from part (iii). The change of variable in the distributional derivative pairing gives $d_n^{\prime }=r_n$; these are the Rademacher functions. Uniform convergence implies weak convergence of distributions, and the continuity of distributional differentiation allows termwise differentiation in that topology. Explicitly,This series assertion is a distributional convergence statement. Its remainder satisfies the concrete estimateIt supplies a well-defined distributional derivative even though the ordinary derivative exists nowhere.

On each interval $[k,k+1]$, $k\in \mathbb{Z}$, the function is $d(x)=x-k$ on $[k,k+\frac{1}{2}]$ and $d(x)=k+1-x$ on $[k+\frac{1}{2},k+1]$. Its ordinary derivative exists off $\frac{1}{2}\mathbb{Z}$ and is the locally integrable functionAssign arbitrary values at the endpoints. For a test function $\varphi$, split $-\int d{\varphi }^{\prime }$ at these corners and apply integration by parts on each piece. The adjacent boundary terms cancel because $d$ is continuous, and only finitely many pieces meet the compact support of $\varphi$. HenceThe distributional derivative is exactly the regular distribution of the ordinary derivative defined almost everywhere. There are no Dirac delta distributions in this first distributional derivative: corners change slope, while a jump of the function itself would create a point mass.

The distributional derivative of $u$ is defined byIt is a distribution because $\varphi \mapsto {\varphi }^{\prime }$ is a continuous map of the space of test functions into itself. If $f\in C^1(a,b)$ represents a regular distribution, integration by parts givesThe boundary term vanishes because $\varphi$ has compact support. Thus the distributional derivative agrees with the ordinary derivative.

For $\mathrm{supp}\varphi \subseteq K$, the assumed order of a distribution estimate givesConsequently $\boxed{\mathrm{ord}(u^{\prime })\le m+1}$.

A distribution on $(a,b)$ is a continuous linear functional on the space of test functions $\mathcal{D}(a,b)=C_c^{\infty }(a,b)$. It has order of a distribution at most $m$ if, for every compact set $K\subset (a,b)$, there is $C_K$ such thatThe order of a distribution is the least such $m$. When saying “of order $m$” means merely that this estimate holds, the wording allows a smaller least order.

For each such $K$, the cutoff function theorem supplies a nonnegative $\chi \in \mathcal{D}(a,b)$ equal to one near $K$. If $M=\Vert \varphi {\Vert }_{\infty }$ and $\mathrm{supp}\varphi \subseteq K$, then $M\chi \pm \varphi \ge 0$ everywhere. Positivity of the linear functional $V$ therefore givesHere $V(\chi )\ge 0$ is finite and independent of $\varphi$. This is precisely the local bound for an order-zero distribution, so the given positive distribution is continuous on the space of test functions. The zero distribution also satisfies the estimate.

With the negative coefficient of $y^{\prime \prime }$, the reduced equation is unchanged, but the fast mode now decays away from the right endpoint. Apply the left boundary condition to the outer solution, givingThe boundary layer still has width $O(ϵ)$. Put $X=(1-x)/ϵ$ and $Y(X)=y(1-ϵX)$. The rescaled equation isIts leading inner expansion has $Y_0=A+B{e}^{-X}$. The asymptotic matching condition is $Y_0(\infty )=e^{-1/2}$, while $Y_0(0)=0$. ThusSubtracting the overlap $e^{-1/2}$ in the additive composite expansion givesIts graph starts exponentially close to one, follows the gently decreasing outer solution, and drops from near $e^{-1/2}$ to zero in a right boundary layer of width $O(ϵ)$. The right endpoint value is exactly zero; the left endpoint error is $-e^{-1/2-1/ϵ}$. Its right endpoint slope is $-e^{-1/2}(1+1/ϵ)$, showing the rapid final descent.

For the positive coefficient of $y^{\prime \prime }$, setting $ϵ=0$ gives the reduced equation $y_0^{\prime }+x{y}_0=0$, with general solution $y_0=C{e}^{-x^2/2}$. A fast decaying homogeneous solution of $ϵy^{\prime \prime }+y^{\prime }=0$ is $e^{-x/ϵ}$, placing the boundary layer at the left endpoint. Thus use the right boundary condition for the outer solution: $y_0(1)=0$ impliesThe dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives width $\delta =ϵ$. With the inner variable $X=x/ϵ$ and $Y(X)=y(ϵX)$, the exact rescaled equation isAt leading order, $Y_{0,XX}+Y_{0,X}=0$, so $Y_0=A+B{e}^{-X}$. The asymptotic matching condition $Y_0(\infty )=y_{\mathrm{out},0}(0)=0$ sets $A=0$, and $Y_0(0)=1$ sets $B=1$. ThereforeThe additive composite expansion has zero overlap to subtract. Its graph starts at one, decreases steeply over $x=O(ϵ)$, and then stays close to zero; its initial slope is $-1/ϵ$. The right endpoint discrepancy $e^{-1/ϵ}$ is exponentially small. The left panel of the figure below sketches this profile. A right-end inner expansion would instead contain a growing $e^X$ term and could not repair the left endpoint mismatch.

Take $ϵ\to 0^{+}$. There are two real branches, on different scales. For a root tending to zero, the Taylor series givesThe dominant balance is $x^3\sim -6ϵ$, so a fractional-power root expansion yieldsMore precisely $x_{-}=-(6ϵ{)}^{1/3}+O(ϵ)$. An ordinary power series in integer powers of $ϵ$ would miss this scale.

There is also a root escaping to positive infinity. Boundedness of the sine term makes the leading dominant balance $x-ϵx^2=0$. The nonzero branch isIndeed the equation rearranges to $ϵx_{+}^2-x_{+}=ϵ-\sin x_{+}$, which yields $x_{+}={ϵ}^{-1}+O(1)$ on this branch. The bounded oscillatory correction need not have a constant limiting value.

To check that these are all the real solutions, solve for the parameter:Positive values occur only on $(-1,0)$ and $(1,\infty )$; $x=\pm 1$ are not roots. A useful exact derivative isIt vanishes only at isolated points, so $E$ is strictly decreasing on each of these intervals. Its endpoint limits are $+\infty$ and zero on each interval. This monotone parametrization of sine perturbation roots gives exactly one root in each interval for every $ϵ>0$, proving that both asymptotic branches, and no others, have been included. If negative small parameters are allowed, the transformation $(x,ϵ)\mapsto (-x,-ϵ)$ reflects these two branches and gives all real solutions for that sign as well.

Consider real roots. Every zero at $ϵ=0$ is $k\pi$, $k\in \mathbb{Z}$, and its derivative $\cos (k\pi )=(-1{)}^k$ is nonzero. For a regular perturbation, write $x=k\pi +\delta$. The Taylor seriesgives $\delta =(-1{)}^kϵ+O({ϵ}^3)$. Thus, through the requested order,There is no quadratic term. The exact parametrization $x_k=k\pi +(-1{)}^k\arcsin ϵ$ also verifies completeness and the remainder order for $|ϵ|<1$.

First take the usual endpoint-regular interpretation $f\in C([-1,1])$. Integrating the divergence form shows that $F=(1-x^2)y^{\prime }$ has finite endpoint limits. A nonzero limit would make $y^{\prime }$ proportional to $1/(1\mp x)$ near that endpoint, forcing a logarithmically unbounded $y$. Thus boundedness requires $F(-1)=F(1)=0$, and consequentlyThis is orthogonality to the constant eigenfunction $P_0$, hence the Fredholm alternative condition. It is also sufficient under the stated endpoint regularity: putIf $M=\max |f|$, the zero-mean condition gives $|F(s)|\le M\min (1+s,1-s)$. Hence the last integrand is bounded, so $y$ extends continuously to both endpoints. Any two bounded solutions differ by a constant: their difference solves $((1-x^2)z^{\prime }{)}^{\prime }=0$, whose other solution is $\frac{1}{2}\log ((1+x)/(1-x))$.

For the requested eigenfunction expansion, setThe Legendre polynomials span all polynomials, so their completeness follows from the Weierstrass approximation theorem. Projecting $\mathcal{L}y=f$ onto each eigenfunction gives $-k(k+1)y_k=f_k$ for $k\ge 1$. The zero mode is free, while $f_0=0$. Thus the full family isThe formula is unchanged by rescaling any $P_k$. For the usual normalization $P_k(1)=1$, the facts $|P_k(x)|\le 1$ on $[-1,1]$ and $N_k=2/(2k+1)$, together with the Cauchy-Schwarz inequality, bound each summand by a constant times $k^{-3/2}$. The series therefore converges absolutely and uniformly. Its identification with the preceding integral solution avoids unjustified termwise second differentiation.

There is a source ambiguity: continuity only on $(-1,1)$ does not supply the endpoint control used above. Even for integrable forcing, zero mean alone is insufficient. For example, letThis $f$ is continuous in the interior, absolutely integrable, and has zero mean because $F(\pm 1)=0$. Yet every candidate with zero endpoint values of $(1-x^2)y^{\prime }$ haswhose integral diverges logarithmically in $H$ at each endpoint. Thus no bounded solution exists. For a merely interior-continuous $f$, the exact criterion is that some constant $B$ makesbounded as $x$ approaches both endpoints. The Bounded solvability for the Legendre operator criterion above covers the intended regular forcing and makes this endpoint limitation explicit.

Write the operator in divergence form:The Legendre differential equation therefore givesThese are the eigenvalues of $\mathcal{L}$. With the common positive-sign convention for a Sturm-Liouville problem, use $-\mathcal{L}$ instead:This is a singular Sturm-Liouville problem, because $p$ vanishes at the endpoints. Bounded eigenfunctions select the nonlogarithmic branches and satisfy $p{P}_n^{\prime }\to 0$ there. The Lagrange identity for a Sturm-Liouville operator givesThe eigenvalues are distinct, proving the orthogonality conditionThe weight is one. No choice of normalization of the Legendre polynomials is needed for this conclusion.

Use $t=x+1$. The Legendre differential equation becomesSubstituting a Frobenius series $y=t^r{\sum }_{k\ge 0}{a}_k{t}^k$, $a_0\ne 0$, gives the indicial equation $2r^2=0$: the indicial root zero is repeated. For $r=0$, the coefficient of $t^k$ yieldsorFor $0\le k<n$ the multiplying factor is nonzero. At $k=n$ it vanishes, so $a_{n+1}$ and all later coefficients are zero. Thus one solution is a polynomial of exact polynomial degree $n$, including a nonzero constant when $n=0$. Denote a chosen normalization by $P_n$.

The linear recurrence relation determines every analytic coefficient from $a_0$, so there cannot be a second independent analytic solution at $t=0$. To see the required logarithm explicitly, reduction of order gives the second solution locally asNear $t=0$, the integrand is $1/(2a_0^{2}t)$ plus an analytic function. After rescaling the second solution, its form is thereforeThe analytic remainder converges locally; its constant term may be set to zero by adding a multiple of $P_n$. The nonzero logarithmic coefficient makes $y_2$ linearly independent of $P_n$. This Frobenius expansion of the Legendre equation at an endpoint gives a Logarithmic Frobenius solution; the same endpoint calculation at $x=1$ explains why the independent Legendre function of the second kind is excluded when boundedness at both ends is required.

Put $N=n(n+1)$. After division by $1-x^2$, the Legendre differential equation has coefficientsBoth are analytic functions at zero, so $\boxed{x=0\text{ is an ordinary point}}$. At either $x_0=\pm 1$, $p$ has a genuine simple pole; hence these are singular points even when $n=0$ makes $q$ vanish. However,and both products extend analytically across $x_0$. By the regular singular point criterion for a second-order equation,None of the three specified points is an irregular singular point.

The nonzero endpoint data contribute to the Fredholm alternative condition. With $w=e^x\sin x$, the same integration by parts identity givesFor the specified forcing the left side is $A{\int }_0^{\pi }{\sin }^{2}xdx=A\pi /2$. HenceTo verify sufficiency, set $y=e^{-x}u$. Direct differentiation gives $\mathcal{L}y=e^{-x}(u^{\prime \prime }+u)$. At this value of $A$, a particular solution of $u^{\prime \prime }+u=(2/\pi )\sin x$ is $-(x/\pi )\cos x$. The complete family obeying the two boundary conditions isThe arbitrary constant records the one-dimensional kernel; existence does not imply uniqueness here.

At $a=\pi$, the nonzero homogeneous solution $e^{-x}\sin x$ satisfies both zero boundary conditions. Thus the boundary value problem has a nontrivial kernel and cannot have an inverse for arbitrary forcing.

More directly, the adjoint operator has the zero-endpoint homogeneous solution $w(x)=e^x\sin x$. If an ordinary Green's function existed, the integration by parts identity would implyBut $w(\xi )\ne 0$ for $0<\xi <\pi$. This contradiction rules out the Green's function for the unmodified operator. It is the Fredholm alternative obstruction.

The homogeneous solution $h(x)=e^{-x}\cos x$ already has the prescribed endpoint values. Thus $z=y-h$ has zero boundary conditions and satisfies $\mathcal{L}z=f$. The Green's function givesEquivalently, splitting the integral at $\xi =x$ gives the explicit formulaAny difference between two solutions would be $e^{-x}(C\cos x+D\sin x)$ with zero endpoint values, forcing $C=D=0$. The displayed solution is therefore unique.

Set $a=\pi /2$. To construct the Green's function from a Dirac delta function, require ${\mathcal{L}}_{x}g(x,\xi )=\delta (x-\xi )$ and zero endpoint values. Away from $x=\xi$, the homogeneous solutions satisfying the left and right boundary conditions are respectivelyWrite $g=A(\xi )u(x)$ for $x<\xi$ and $g=B(\xi )v(x)$ for $x>\xi$. A discontinuity in $g$ would generate a derivative of the Dirac delta function, which is absent from the desired equation. Thus $g$ is continuous. Integrating that equation through $\xi$ then gives the jump conditionThe resulting two equations are $Au(\xi )=Bv(\xi )$ and $B{v}^{\prime }(\xi )-A{u}^{\prime }(\xi )=1$. Since the Wronskian $u{v}^{\prime }-u^{\prime }v=-e^{-2\xi }$, they giveBoth endpoint values vanish. At $x=\xi$, subtracting the two first derivatives gives ${\sin }^2\xi +{\cos }^2\xi =1$, verifying the sign and normalization of the Dirac delta function.