Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 5 / 1

The Taylor series definition says that $f\in C^{\infty }(\mathbb{R})$ and, for every $a$, there is $r>0$ such thatThe equivalent factorial derivative criterion for real analyticity says that, for every $a$, there are a neighborhood $I$ of $a$ and constants $C,A>0$ such thatThe uniformity over $I$ matters: bounds only at $a$ do not exclude a flat function.

Assume the factorial derivative criterion for real analyticity. The Taylor theorem with Lagrange remainder giveswhen the segment from $a$ to $a+h$ is contained in $I$. For sufficiently small $|h|<A^{-1}$ the Taylor remainder tends to zero, proving the Taylor series definition.

Conversely, write the convergent power series at $a$ as $\sum b_k{h}^k$. Choose $\rho$ strictly inside its radius of convergence; then $|b_k|\le M{\rho }^{-k}$ for some $M$. Termwise differentiation on $|h|\le \rho /2$ givesHere the sum is $(1-1/2{)}^{-n-1}$, obtained by differentiating the geometric series. This is the required locally uniform bound. The two definitions of a real analytic function are equivalent.

Consider the flat functionAway from zero every derivative has the form $P_n(1/x)e^{-1/x^2}$ for a polynomial $P_n$: differentiating preserves this form. For every $m\ge 0$,because an exponential function decays faster than any power. Inductively, extend each displayed derivative by zero at zero. It is continuous there, and its difference quotient at zero also tends to zero by the same estimate with one extra power of $|x{|}^{-1}$. Thus each extension is the derivative of the preceding extension. This proves $f\in C^{\infty }(\mathbb{R})$ and $f^{(n)}(0)=0$ for all $n$.

Its Taylor series at zero is identically zero, whereas $f(x)>0$ for every $x\ne 0$. It is smooth everywhere but not real analytic at zero.

The complex Liouville theorem states that a bounded entire function is constant. Indeed, if $|f|\le M$, the Cauchy estimate on any disc of radius $R$ centered at $z$ gives $|f^{\prime }(z)|\le M/R$. Letting $R\to \infty$ gives $f^{\prime }(z)=0$ everywhere.

The analogous conclusion for bounded real analytic functions on $\mathbb{R}$ is false. For example, $\sin x$ is bounded, nonconstant, and real analytic on the whole real line. Boundedness only on that line does not bound its holomorphic extension on the complex plane.

A separable Hilbert space has a countable subset dense in its norm topology.

The printed assertion about all locally square-integrable functions is false. The proposed average is not even finite for every such function: for $f(x)=x$,It also fails positive definiteness. The nonzero function $f=1_{[0,1]}$ hasConsequently this formula cannot define an inner product, much less a Hilbert space, on $L_{\mathrm{loc}}^2(\mathbb{R})$.

A precise version of the intended nonseparability argument uses the mean-square completion of trigonometric polynomials. Start with the real vector space $V$ of finite linear combinations of $1$, $\cos (\lambda x)$, and $\sin (\lambda x)$, with arbitrary $\lambda >0$. Product-to-sum identities show that all the proposed cross averages exist. Distinct frequencies are orthogonal, each sine and cosine has squared norm one, and the constant function has squared norm two. Thus, after collecting equal frequencies,This is positive definite on $V$. Its Hilbert space completion $H$ contains the uncountable orthonormal set $\{\cos (\lambda x):\lambda >0\}$. The distance between two distinct members is $\sqrt{2}$. Their open balls of radius $1/2$ are pairwise disjoint, and a dense subset must meet each one. A countable dense subset is therefore impossible: this corrected completed space is nonseparable. Completion is an essential additional construction; it does not validate the printed claim about all of $L_{\mathrm{loc}}^2$.

The closest point theorem in a Hilbert space says that, for every nonempty closed convex set $C\subset H$ and $x\in H$, there is exactly one $p\in C$ minimizing $\Vert x-p\Vert$.

Put $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The midpoint belongs to $C$ because it is a convex set. The parallelogram law givesThus $(y_n)$ is a Cauchy sequence. Completeness of the Hilbert space and closedness of $C$ give a limit $p\in C$, with $\Vert x-p\Vert =d$. Applying the same identity to two minimizers gives their squared distance at most zero, proving uniqueness.

The resulting projection is characterized byIndeed, differentiate $\Vert x-p-t(z-p){\Vert }^2$ at $t=0^{+}$; the minimum there gives the inequality. Conversely, expanding $\Vert x-z{\Vert }^2$ proves minimality from this inequality. For a closed linear subspace, both signs of each direction are allowed, so $x-p$ is orthogonal to that subspace: this recovers the orthogonal projection.

The Riesz representation theorem states that every bounded linear functional $L$ on a real or complex Hilbert space $H$ is represented by a unique $h\in H$:For the complex case take the inner product to be linear in its first argument.

If $L=0$, choose $h=0$. Otherwise its kernel $K$ is a closed linear subspace. Choose $x$ with $L(x)\ne 0$ and let $z=x-P_{K}x$, using the orthogonal projection. Then $z\ne 0$, $z\perp K$, and $L(z)=L(x)\ne 0$. For every $v$,Therefore take $h=L(z)z/\Vert z{\Vert }^2$ in the real case, and $h=\overline{L(z)}z/\Vert z{\Vert }^2$ in the complex case. The conjugate in the latter formula compensates for conjugate linearity in the second argument.

The Cauchy-Schwarz inequality gives $|L(v)|\le \Vert v\Vert \Vert h\Vert$, and evaluation at $v=h/\Vert h\Vert$ when $h\ne 0$ gives equality of the norms. If two vectors represent $L$, their difference is orthogonal to every vector, including itself, hence zero. This proves all assertions of the Riesz representation theorem.

The real Lax-Milgram theorem applies to a Hilbert space $H$ and a bounded bilinear form $a$ satisfyingFor every bounded linear functional $L$ there is a unique $u\in H$ withSymmetry of the bilinear form is not required.

By the Riesz representation theorem, write $a(u,v)=⟨Au,v⟩$ and $L(v)=⟨g,v⟩$. The operator $A$ is linear and bounded, with $\Vert A\Vert \le M$. The coercive bilinear form bound and the Cauchy-Schwarz inequality implyHence $A$ is injective. Its range is closed: if $A{u}_n$ converges, this last inequality applied to differences makes $u_n$ a Cauchy sequence, and its limit maps to the proposed range limit. If $w$ is in the orthogonal complement of the range, then $a(u,w)=0$ for all $u$; taking $u=w$ and using coercivity gives $w=0$. The range is thus dense as well as closed, so it is all of $H$. Solve $Au=g$ uniquely; the displayed lower bound gives the asserted estimate.

For complex Hilbert spaces the same proof works for a bounded sesquilinear form, linear in the first argument, with $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }^2$. In the convention $a(u,v)=L(v)$, $L$ must then be a bounded conjugate-linear functional represented as $⟨g,v⟩$.