For a fixed centre $a$, the triangle inequality gives $|d(a,x)-d(a,y)|\le d(x,y)$, so the distance-to-centre function is continuous. Its preimage of $(-\infty ,r]$ is the closed ball, hence is closed. Since that closed set contains the corresponding open ball, it contains its closure:Here the bar on the right denotes a closed ball, while the bar on the left denotes closure.

On a product of sigma-finite measure spaces, Tonelli's theorem equates product and iterated integrals for a nonnegative measurable function, allowing infinite values. Fubini's theorem says that an absolutely integrable measurable function has integrable sections almost everywhere, their section integrals are integrable, and both iterated integrals equal the product integral. Absolute integrability may be checked first by applying Tonelli to the absolute value.

The partial sums $S_N={\sum }_{k=0}^N{a}_k$ are measurable, nonnegative and increasing. By the monotone convergence theorem,A nonnegative measurable function with finite integral is finite almost everywhere: if it were infinite on a positive-measure set, every constant multiple of that set's indicator would be a lower bound, forcing an infinite integral. Thus the series converges almost everywhere.

An invertible matrix is an ordered basis of $K^n$. After choosing $j$ independent columns, their span has $q^j$ elements, so the next column has $q^n-q^j$ choices. Thus

Starting with $A_0=A$, choose real shifts ${\sigma }_j$, factor $A_j-{\sigma }_{j}I=Q_j{R}_j$, and setEach step is an orthogonal similarity, so the eigenvalues are preserved and all iterates are similar. If $A_j^T=A_j$, transposing $Q_j^T{A}_j{Q}_j$ proves symmetry of $A_{j+1}$, completing the induction.

The given matrix is the Householder reflectionIt has eigenvalue $-1$ along $1$ and $+1$ on its orthogonal complement, hence $A^{T}A=I$. In the positive-diagonal QR convention its unshifted factorization is $Q=A$, $R=I$, so the next iterate is again $A$. Thus unshifted QR is stationary, despite the matrix generally having nonzero off-diagonal entries. If arbitrary diagonal signs are allowed in $R$, a step can instead replace $A$ by $DAD$ for a diagonal sign matrix $D$; every off-diagonal magnitude remains unchanged. Such sign changes still make no progress towards diagonalization.

Only the first and third coordinates occur in the first two columns. Normalize the first column to get $q_1=(4,0,3,0{)}^T/5$. Removing its component from the second column leaves $(21,0,-28,0{)}^T/5$, of length seven, so $q_2=(3,0,-4,0{)}^T/5$. The remaining orthonormal columns can be $q_3=e_2$ and $q_4=e_4$. Taking their dot products with the columns of $A$ givesThe columns of $Q$ are orthonormal, and direct multiplication gives the source matrix. Exploiting its two-coordinate first block avoids unnecessary four-dimensional elimination.

A QR factorization is $A=QR$, where $Q^{T}Q=I$ and $R$ is upper triangular. If $A$ is orthogonal, so is $R=Q^{T}A$. Its inverse is upper triangular, but $R^{-1}=R^T$ is lower triangular, forcing $R$ to be diagonal. Orthogonality then givesWith the customary positive diagonal convention, $R=I$.

The residual is orthogonal to $q\in {\Pi }_k$, so Pythagoras givesTaking nonnegative square roots proves $\boxed{\Vert q\Vert \le \Vert f\Vert }$.

The singular weight requires care about the space of admissible polynomials. Since $n\ge 1$, $x^n$ has finite squared norm. A polynomial $q$ has finite error relative to $x^n$ near zero exactly when $q(0)=0$: a nonzero constant term makes the integrand asymptotic to $q(0{)}^2/x$, whose integral diverges. If $q(0)=0$, the residual is $O(x)$ and the integrand is $O(x)$.

Thus the full space ${\Pi }_{n-1}$ is not a subspace of $V$, so the projection theorem cannot be applied to it as written. Nevertheless a unique best finite-error approximation does exist inFor $n=1$, this is the zero space and the answer is $q=0$. For $n\ge 2$, $W$ is finite dimensional and its Gram matrix is positive definite, because a nonzero polynomial cannot vanish on an interval. Solving the normal equations therefore gives its unique orthogonal projection of $x^n$. All omitted polynomials have infinite error, so this also minimizes the extended error over the full polynomial set. The distinction is an invalid full-space projection premise, not nonexistence of every finite-error best approximation. As throughout weighted $L^2$, use equivalence classes for the underlying measure.

Write $q=A+Bx$ and $\ell =\log 3$. Orthogonality of $x^2-q$ to $1$ and $x$ givesThe Gram matrix is positive definite, with determinant $4(\ell -1)>0$, so the solution is unique. Solving yields

Use the usual weighted $L^2$ interpretation, identifying functions equal almost everywhere; on raw pointwise functions the displayed integral is only a seminorm. Assume $q\in {\Pi }_k\subset V$. If $f-q$ is orthogonal to ${\Pi }_k$, then for every $p\in {\Pi }_k$,Conversely, suppose $q$ minimizes the error. For $r\in {\Pi }_k$ and every real $t$, compare $q$ with $q+tr$:If the linear coefficient were nonzero, a sufficiently small $t$ with the appropriate sign would make the expression negative. Thus $⟨f-q,r⟩=0$ for every $r$, proving best approximation is equivalent to orthogonality of the residual.