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Fix an integer $k\ge 1$ and consider the independent incrementsFor fixed $k$ these are independent and identically distributed random variables with the normal distribution $N(0,1/k)$, so $p_k=\mathbb{P}(|{\Delta }_{j,k}|>1)>0$. The second Borel-Cantelli lemma gives $|{\Delta }_{j,k}|>1$ for infinitely many $j$, almost surely. Intersecting these probability-one events over the countably many $k$ preserves probability one.

On any path in this intersection, uniform continuity on $[0,\infty )$ would give some $\delta >0$ such that $|B_t-B_s|<1$ whenever $|t-s|<\delta$. Choosing $k$ with $1/k<\delta$ contradicts the existence of an increment $|{\Delta }_{j,k}|>1$. Therefore Brownian paths are not uniformly continuous on the half-line:The event is measurable: for continuous paths, uniform continuity can be tested by countably many pairs of nonnegative rational numbers and tolerances $1/m$. Brownian motion has uniform continuity on every compact time interval, but almost surely fails it on the whole half-line.

For the dyadic power variation of a continuous local martingale, write ${\Delta }_i^{n}M=M_{i{2}^{-n}}-M_{(i-1)2^{-n}}$, ${\delta }_n={\max }_i|{\Delta }_i^{n}M|$ and $Q_n={\sum }_i({\Delta }_i^{n}M{)}^2$. Uniform continuity on $[0,1]$ gives ${\delta }_n\to 0$ almost surely, while the deterministic-partition characterization of quadratic variation gives $Q_n\to ⟨M{⟩}_1$ in probability. In particular $(Q_n)$ is bounded in probability.

If $p>2$, then $Z_n\le {\delta }_n^{p-2}{Q}_n$, and a factor tending to zero in probability times a sequence bounded in probability tends to zero in probability. ThusIf $1<p<2$, the reverse comparison is $Q_n\le {\delta }_n^{2-p}{Z}_n$. The assumed finite $\mathrm{limsup}{Z}_n$ makes $Z_n$ eventually bounded almost surely, so $Q_n\to 0$ almost surely. By uniqueness of a limit in probability, $⟨M{⟩}_1=0$ almost surely. Its monotonicity gives $⟨M{⟩}_t=0$ for $t\le 1$.

To finish without assuming any integrability of $M$, apply the bound from part (i) to both $M$ and $-M$, with $t=1$, and then let $b\downarrow 0$. For each $a>0$ the probability of a positive or negative excursion beyond $a$ is zero. Taking countably many $a\downarrow 0$ proves $M$ is indistinguishable from zero on $[0,1]$. The same argument in fact works for every $0<p<2$ under the stated pathwise boundedness assumption.

First address the positivity request in part (a), which has no separate Solution heading in the stub. For $x\in [0,1]$, each $x^k(1-x{)}^{n-k}$ and each coefficient $c_k$ is nonnegative. Their sum is therefore nonnegative, including at the endpoints:This is the elementary direction of the positive Bernstein coefficients for a strictly positive polynomial criterion; the converse developed later needs strict positivity.

For completeness, the convergence fact listed as (i) also has a short proof. If $K\sim \mathrm{Bin}(n,x)$, the Bernstein polynomial is $B_n(f)(x)=\mathbb{E}[f(K/n)]$. Its mean argument is $x$ and $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$. Given $ϵ>0$, uniform continuity of $f$ on the compact interval supplies $\delta >0$ such that $|f(s)-f(t)|<ϵ$ whenever $|s-t|\le \delta$. By Chebyshev's inequality, uniformly in $x$,Letting $n\to \infty$ and then $ϵ\downarrow 0$ proves $\boxed{\Vert B_n(f)-f{\Vert }_{\infty }\to 0}$. The two numbered items in the source are assumptions supplied for later parts, not additional exam questions; their stub sections contain these supporting derivations.

For $f\in C[0,1]$, define its Bernstein polynomialIf $X_{n,x}\sim \mathrm{Bin}(n,x)$, thenLet $M=\Vert f{\Vert }_{\infty }$. By uniform continuity, for any $\epsilon >0$ choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon /2$ whenever $|u-v|<\delta$. Splitting the expectation according to that event and using Chebyshev's inequality gives, uniformly in $x$,For sufficiently large $n$ this is below $\epsilon$. Thus $B_{n}f\to f$ uniformly on $[0,1]$. An affine change of variable treats every compact interval, proving the Weierstrass approximation theorem.

A function $f:A\to \mathbb{R}$ is uniformly continuous if, for every $\epsilon >0$, there is a $\delta >0$ such thatThe same $\delta$ must work at every pair of points in the domain.

Let $C\subset \mathbb{R}$ be compact and let $M=\Vert f{\Vert }_{\infty }$. Since every $k_n$ is supported in $[-R,R]$, the function $f$ is uniformly continuous on the compact setGiven $\epsilon >0$, choose $\delta >0$ so thatwhenever $x\in C$ and $|t|<\delta$. Since $\int k_n=1$ and $k_n\ge 0$,Property 3 makes the final term smaller than $\epsilon /2$ for all sufficiently large $n$, uniformly in $x\in C$. Hence $f_n\to f$ uniformly on every compact set. The sequence $(k_n)$ is an approximate identity.

For the second part, extend $g$ to a continuous function $\tilde{g}$ on $\mathbb{R}$ by setting it equal to zero outside $[0,1]$; continuity at the endpoints uses $g(0)=g(1)=0$. DefineThese nonnegative kernels have integral one. For every $\delta >0$, their mass outside $[-\delta ,\delta ]$ tends to zero exponentially relative to the mass near zero, so they satisfy property 3.

For $x\in [0,1]$, the convolution isBecause $|x-y|\le 1$ on the square $[0,1{]}^2$, no cutoff remains in this formula. Expanding the $n$th power shows that $p_n$ is a polynomial in $x$. The first part, applied to the compact interval $[0,1]$, givesThis proves the Weierstrass approximation theorem for functions with the stated endpoint values.

Let $\Vert K{\Vert }_{\infty }={\max }_{[0,1{]}^2}|K|$. For $f,g\in C([0,1])$,Taking the supremum over $x$ proves that the integral operator $T$ is Lipschitz continuous, hence continuous, in the uniform norm.

It remains to see that $Tf$ is continuous. Since $K$ is uniformly continuous on the compact square and $f$ is bounded,as $x^{\prime }\to x$. Therefore $T:C([0,1])\to C([0,1])$ is well defined and continuous.

Fix $x\in {\mathbb{R}}^n$ and a coordinate direction $e_i$. The fundamental theorem of calculus givesFor sufficiently small $h$, the points $(t,x+sh{e}_i)$ lie in a fixed compact set. The continuity of $D_{i}g$ makes it uniformly continuous there, so the right-hand side converges to $D_{i}g(t,x)$ uniformly in $t$. We may consequently pass the limit through the finite integral:

To prove continuity, if $x_k\to x$, joint continuity makes $D_{i}g(t,x_k)\to D_{i}g(t,x)$ uniformly for $t\in [0,1]$ once the $x_k$ lie in a compact neighbourhood of $x$. Hence $D_{i}G(x_k)\to D_{i}G(x)$. This proves the stated differentiation under the integral sign result and the continuity of every partial derivative.

Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and

This is always true. Since $f$ is bounded and $f_n\to f$ uniformly, there are $M$ and $N$ such thatfor every $x\in X$ and $n\ge N$. The continuous function $g$ is uniformly continuous on the compact interval $[-M-1,M+1]$. Therefore, given $\epsilon >0$, a sufficiently small uniform bound on $|f_n-f|$ impliesHence $g\circ f_n\to g\circ f$ uniformly.

Let $X\sim \mathrm{Bin}(n,t)$ and $Y\sim \mathrm{Bin}(n,s)$ be independent. The displayed Bernstein polynomial isGiven $\epsilon >0$, uniform continuity of $f$ supplies $\delta >0$ such that changing both coordinates by less than $\delta$ changes $f$ by less than $\epsilon$. If $M=\Vert f{\Vert }_{\infty }$, then Chebyshev inequality and
$\mathrm{Var}(X/n),\mathrm{Var}(Y/n)\le 1/(4n)$ give, uniformly in $(t,s)$,Thus $B_{n}f\to f$ uniformly.

Each $B_{n}f$ is a polynomial, whose iterated integrals may be interchanged term by term. Uniform convergence permits passage to the limit in both iterated integrals, proving the asserted continuous-function form of Fubini's theorem.

Now suppose every monomial moment of $f$ vanishes. By linearity,
$\int \int pf=0$ for every two-variable polynomial $p$. The just-proved Bernstein approximation gives polynomials $p_n\to f$ uniformly. Thereforeso $\boxed{f=0}$. Statement (i) is true.

Let ${\rho }_{n,j}^{\prime }$ be the $j$th one-system marginal of ${\rho }_n^{\prime }$, and put ${\epsilon }_n=\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1$. Contractivity of trace distance under partial trace givesuniformly in $j$. Because the one-system state space is a compact space, continuity of $f(\cdot \Vert \sigma )$ implies uniform continuity. Hence there is a function $\delta (\epsilon )\to 0$ such thatfor every $j$.

Multipartite superadditivity and additivity now implyThereforewhich proves the required lower asymptotic semicontinuity of quantum relative entropy argument.

A product of two uniformly continuous functions need not have uniform continuity on an unbounded domain. The functions $f(x)=g(x)=x$ give the counterexample $x^2$, since inputs $n$ and $n+1/n$ approach each other while their squared values remain separated. For bounded factors the estimate $|f(x)g(x)-f(y)g(y)|\le \Vert f{\Vert }_{\infty }|g(x)-g(y)|+\Vert g{\Vert }_{\infty }|f(x)-f(y)|$ restores uniform continuity.

For $f\in L^1(\mathbb{T},m)$ with $m$ normalized Lebesgue measure, translation on the circle group is continuous in the L1 norm: $\Vert f(\cdot +t)-f{\Vert }_1\to 0$ as $t\to 0$ modulo one. Approximate $f$ in the L1 norm by a continuous function, use its uniform continuity, and use translation invariance to preserve the approximation error. In particular, $m(B△(B-t))\to 0$ for every measurable $B$.

Choose the uniform continuity input tolerance corresponding to output tolerance one. A totally bounded domain is covered by finitely many balls of smaller radius centred in the domain. Every function value differs by less than one from the value at one of their finitely many centres. Taking the largest centre value bounds the whole function. Compactness or closedness of the domain is unnecessary.

If a uniformly continuous function $u:{\mathbb{R}}^d\to \mathbb{R}$ belongs to Lp space for some $1\le p<\infty$, then $u(x)\to 0$ as $|x|\to \infty$. Otherwise choose points escaping to infinity with $|u(x_j)|\ge 2\theta >0$. Uniform continuity gives one radius $\rho >0$ on which $|u|\ge \theta$ around every $x_j$. Extract disjoint such balls. Their contributions to $\int |u{|}^p$ are each at least ${\theta }^p|B_{\rho }|$, a contradiction. The finite-$p$ condition is essential: the constant function one lies in $W^{1,\infty }$ and does not vanish at infinity.

If $\Theta$ is compact, $q(\theta ,x)$ is continuous in $\theta$ and measurable in $x$, and $\mathbb{E}{\sup }_{\Theta }|q(\theta ,X)|<\infty$, then an independent and identically distributed sample satisfies ${\sup }_{\Theta }|P_n{q}_{\theta }-P{q}_{\theta }|\to 0$ almost surely. Uniform continuity for each $x$, the dominated convergence theorem, and finite parameter nets produce arbitrarily narrow function brackets.