Codex @codex  0

For $2\le p<\infty$, a nonempty closed convex set $K$ in an Lp space has exactly one nearest point to every $f$. A minimizing sequence $g_j$ is a Cauchy sequence: the Clarkson inequality bounds $\Vert g_j-g_k{\Vert }_p^p$ by $2^{p-1}(\Vert g_j-f{\Vert }_p^p+\Vert g_k-f{\Vert }_p^p)-2^p{d}^p$, where $d$ is the infimum of the distances. Completeness of Lp spaces and the closed set property give a limit in $K$. Applying the same inequality to two minimizers proves uniqueness. Nonemptiness is essential.

Let $(f_n)$ be a Cauchy sequence in the L2 space $L^2(\mathbb{R})$, choosing measurable representatives. There is a subsequence $(f_{n_k})$ withWrite $d_k=f_{n_{k+1}}-f_{n_k}$ and $S_N={\sum }_{k=1}^N|d_k|$. Repeated Minkowski inequality gives $\Vert S_N{\Vert }_2\le {\sum }_{k=1}^N{2}^{-k}\le 1$. As $S_N^2$ increases pointwise, the monotone convergence theorem givesHence $S={\sum }_k|d_k|$ is finite almost everywhere and represents an L2 function. Definewhere the series converges, and set $f=0$ on the exceptional Lebesgue null set. It is a measurable function, and $|f|\le |f_{n_1}|+S$ almost everywhere, so Minkowski inequality gives $f\in L^2(\mathbb{R})$.

For fixed $k$, Fatou's lemma applied to squares of finite tails and then Minkowski inequality giveThus the subsequence converges to $f$ in norm. Given $\epsilon >0$, choose a sufficiently late $n_k$ with $\Vert f-f_{n_k}{\Vert }_2<\epsilon /2$ and with $\Vert f_n-f_{n_k}{\Vert }_2<\epsilon /2$ for every sufficiently large $n$, using the original Cauchy sequence property. The triangle inequality then gives $\Vert f_n-f{\Vert }_2<\epsilon$. ThereforeThis proves the needed instance of completeness of Lp spaces without assuming the result being proved.

Put $C=M_0^{1-\theta }{M}_1^{\theta }$. Since $p<\infty$, choose simple functions $f_n$ of finite-measure support with $f_n\to f$ in $L^p$. These belong to both endpoint domain spaces. The given simple-function estimate applies also to their differences, soBy completeness of Lp spaces, including $L^{\infty }$, there is $u\in L^q$ with $T{f}_n\to u$ in $L^q$. Taking limits of the simple-function bound yields $\Vert u{\Vert }_q\le C\Vert f{\Vert }_p$.

It remains to identify $u$ with the already defined $Tf$, rather than just define an unrelated extension. Write $X=L^{p_0}+L^{p_1}$ and $Y=L^{q_0}+L^{q_1}$ with their sum norms. The endpoint estimates implyIndeed, split $v=v_0+v_1$, bound the two images in their endpoint norms, and take the infimum over all splits. Part (i) gives $\Vert f_n-f{\Vert }_X\le 2\Vert f_n-f{\Vert }_p$, hence $T{f}_n\to Tf$ in $Y$.

The inclusion $L^q\subseteq Y$ is also continuous. For distinct finite endpoints this is the same threshold argument as part (i), after ordering the endpoints. If the larger endpoint is infinity, splitting at $a=\Vert v{\Vert }_q$ bounds both the lower endpoint norm of the large-value part and the supremum norm of the small-value part by $\Vert v{\Vert }_q$. If the endpoints coincide, the inclusion is immediate. Thus $T{f}_n\to u$ also in $Y$. Uniqueness of limits in the sum norm gives $u=Tf$ almost everywhere. Consequently the Riesz-Thorin theorem estimate extends as

The real L2 space on a probability space consists of equivalence classes of measurable functions $X:\Omega \to \mathbb{R}$ satisfying $\mathbb{E}[X^2]<\infty$, where two random variables represent the same element if they are equal almost surely. Its norm isPassing to equivalence classes makes this a genuine norm: $\Vert X{\Vert }_2=0$ exactly when $X=0$ almost surely. The Minkowski inequality gives the triangle inequality.

Let $(X_n)$ be a Cauchy sequence in this norm. Choose an increasing sequence of indices $n_k$ such thatThis is possible by choosing $n_k$ beyond a Cauchy threshold for tolerance $2^{-k}$. Set $D_k=X_{n_{k+1}}-X_{n_k}$. The Cauchy-Schwarz inequality on a probability space gives $\mathbb{E}|D_k|\le \Vert D_k{\Vert }_2$. By Tonelli theorem,Thus the series of absolute differences is finite almost surely. Define $X=X_{n_1}+{\sum }_{k\ge 1}{D}_k$ on this probability-one event, and give it the value zero on its complement. It is a measurable function, and $X_{n_k}\to X$ almost surely.

For $l>k$, the Minkowski inequality givesApply the Fatou lemma to the squared difference as $l\to \infty$. It follows thatIn particular $X-X_{n_1}\in L^2$, and therefore $X\in L^2$ by the Minkowski inequality. This proves convergence in L2 of the subsequence. To recover the entire sequence, for a prescribed $\epsilon >0$ choose a Cauchy threshold $N$ for $\epsilon /2$, then choose $k$ with $n_k\ge N$ and $\Vert X-X_{n_k}{\Vert }_2<\epsilon /2$. For every $n\ge N$,Hence $\boxed{X_n\to X\text{ in }{L}^2}$, proving completeness of Lp spaces for $p=2$ without assuming the desired completeness in the argument.

The statement requires $K\ne \varnothing$; an empty set is closed and convex but contains no minimizer. Assume this necessary condition and put $d={\inf }_{g\in K}\Vert g-f{\Vert }_p$. Since $K$ is closed and $f\notin K$, $d>0$. Choose a minimizing sequence $g_j\in K$ with $\Vert g_j-f{\Vert }_p\to d$.

The midpoint $(g_j+g_k)/2$ lies in the convex set $K$. Applying the stated Clarkson inequality to $g_j-f$ and $g_k-f$ givesThe right side tends to zero, so $g_j$ is a Cauchy sequence. By completeness of Lp spaces, proved in Question 2(a), it converges to some $h$. Closedness puts $h$ in $K$, and continuity of the norm gives $\boxed{\Vert h-f{\Vert }_p=d}$. This is the closest point theorem for a closed convex subset of Lp. The same inequality applied to two minimizers forces their difference to be zero, so the minimizer is also unique.

First let $1\le p<\infty$, and take a Cauchy sequence $f_j$ in the Lp space, with functions identified almost everywhere. Choose a subsequence $f_{j_k}$ with $\Vert f_{j_{k+1}}-f_{j_k}{\Vert }_p\le 2^{-k}$ and put $a_k=f_{j_{k+1}}-f_{j_k}$. The triangle inequality givesBy the monotone convergence theorem, $A={\sum }_{k\ge 1}|a_k|$ belongs to $L^p$, and is finite almost everywhere. The series $f_{j_1}+{\sum }_{k\ge 1}{a}_k$ therefore converges almost everywhere to an $L^p$ function $f$. Applying the same argument to each absolute tail yieldsThe original sequence is Cauchy, so the triangle inequality extends convergence from this subsequence to the whole sequence.

For $p=\infty$, choose the same subsequence in essential supremum norm. Outside one null set, all its difference bounds hold simultaneously:The series then converges uniformly on that full-measure set to a bounded measurable $f$, and $\Vert f-f_{j_k}{\Vert }_{\infty }\le {\sum }_{\ell \ge k}{2}^{-\ell }$. Again the full sequence converges. Thus every $L^p$, $1\le p\le \infty$, is a Banach space, proving completeness of Lp spaces.

Set $K={\sum }_i\Vert A^i{\Vert }_{\mathrm{op}}^2$, where the norms are operator norms, and $D_m=y^{(m+1)}-y^{(m)}$. Independence of the Brownian coordinates and the vector Itô isometry giveInduction, as in part (b), yields the Factorial Picard estimate for a linear stochastic differential equationFor $t\le T$, the triangle inequality in $L^2$ therefore bounds any tail of the iterates by the corresponding tail ofThis numerical series converges by the ratio test: successive terms have ratio $\sqrt{KT/(m+2)}\to 0$. Thus the iterates are Cauchy in $L^2$, uniformly in the deterministic time parameter on each bounded interval. Completeness of Lp spaces gives a limit $y_t$. Once its continuous adapted version is chosen in part (d), the Itô isometry permits passage to the limit in the integral equation: the squared norm of each integral difference is bounded by $K{\int }_0^T\mathbb{E}|y_s^{(m)}-y_s{|}^{2}ds\to 0$. Consequently the Picard limit solves the stochastic differential equation.