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Let $M(x)={\sum }_{n\le x}\mu (n)$ be the Mertens function. Partial summation givesFor any $\Re s>0.9$, choose a sufficiently small $ϵ>0$ with $0.9+ϵ<\Re s$. The assumed bound makes the boundary term tend to zero and makes the integral absolutely convergent. Moreover, on each compact set of this half-plane, one can choose the same $ϵ$, so the integral has locally uniform convergence. Henceis a holomorphic function on $\Re s>0.9$, agreeing with the reciprocal Dirichlet series from part (b) on $\Re s>1$.

The identity theorem applied to the holomorphic functions $(s-1)\zeta (s)F(s)$ and $s-1$ extends their equality throughout $\Re s>0.9$; the assumed simple pole makes the first function holomorphic at $1$. Thus $\zeta (s)F(s)=1$ away from $1$, proving a zero-free half-plane:

To obtain the strongest symmetric conclusion, use the following basic facts explicitly: the functional equation of the Riemann zeta function gives the entire function $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ with $\xi (s)=\xi (1-s)$, and the trivial zeros of the Riemann zeta function are the negative even integers. Every Nontrivial zero of the Riemann zeta function is therefore accompanied by its reflection $1-s$. A Nontrivial zero of the Riemann zeta function with real part below $0.1$ would reflect to one above $0.9$, which is impossible. Consequently all nontrivial zeros lie in the closed stripThe $ϵ$ in the hypothesis prevents this argument from excluding either boundary line; the trivial zeros of the Riemann zeta function remain present.

No: local uniform convergence need not be global uniform convergence. For a bounded counterexample use the continuous functions $f_n(x)=\min (1,|x|/n)$ and $f(x)=0$. On any fixed interval $[-M,M]$, for $n>M$,Thus there is locally uniform convergence to zero. On all of $\mathbb{R}$, however, ${\sup }_x|f_n(x)-f(x)|=1$ for every $n$, so uniform convergence fails. The cutoff needed to control the tails cannot be chosen independently of the interval.

Put $M_k=S_k-mk$ as above. Centering the linear interpolation also interpolates the centered values: for $u\in [0,1]$,The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore givesHence there is uniform convergence on compacts in probability to the deterministic path $h(t)=mt$, by Markov inequality. Equip $C([0,\infty ),\mathbb{R})$ with its standard compact-open topology, metrized byFor each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by ${\sum }_{j>J}{2}^{-j}$. Thus $d(S^{(N)},h)\to 0$ in probability. Convergence in probability to a deterministic point implies weak convergence of probability measures, so the fluid limit isHere ${\delta }_h$ is the Dirac measure concentrated on that continuous path. The topology is locally uniform convergence; no assertion of uniform convergence over the entire unbounded half-line is needed. The interpolation is intended for integers $k\ge 0$, including the initial interval $[0,1]$. If the PDF's ${\mathbb{Z}}^{+}$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.

Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

Let the common domain be an open set $U$. The uniform limit theorem makes the limit $f$ continuous. For every triangle whose closed interior lies in $U$, Cauchy's integral theorem gives ${\int }_{\partial T}{f}_n=0$. Part (ii) passes this identity to the limit, giving ${\int }_{\partial T}f=0$. Morera's theorem now proves that $f$ is holomorphic.

The same reasoning works for locally uniform convergence, since each triangle boundary is compact. No passage of derivatives through a limit has been assumed.

The answers areWrite $M_n(f,g)={\sup }_{0\le t\le n}|f(t)-g(t)|$. The given metric describes locally uniform convergence: $d(f_j,f)\to 0$ if and only if $M_n(f_j,f)\to 0$ for every fixed $n$. Indeed,so convergence in $d$ forces $M_n\to 0$. Conversely, split the defining series at $N$: its tail is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.

If $(f_j)$ is a Cauchy sequence in $d$, the same single-term estimate makes it a Cauchy sequence in the supremum norm on every $[0,n]$. The completeness of the continuous-map space in the uniform metric gives a continuous function $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the limit of a sequence gives $F_{n+1}{|}_{[0,n]}=F_n$, so these functions define a single continuous function $f$ on $[0,\infty )$. The preceding finite-head and tail argument proves $d(f_j,f)\to 0$. Thus the space is complete.

The constant functions $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no convergent subsequence, so the space is not sequentially compact.

For fixed $f,g$, set $\gamma (s)=(1-s)f+sg$. On $[0,n]$,as $s^{\prime }\to s$. The equivalence with locally uniform convergence proves that $\gamma$ is a continuous path for $d$. Hence the space is path-connected and connected.

Let $f_n$ be continuously differentiable functions on an interval $I$. If $f_n^{\prime }\to g$ by uniform convergence on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a continuously differentiable function $f$ with $f^{\prime }=g$. If $I$ is bounded, the convergence is uniform on all of $I$.^[ref]

The anchor fixes the additive constant that a derivative cannot see. By the fundamental theorem of calculus, defineThe uniform limit theorem makes $g$ a continuous function, so the fundamental theorem of calculus gives $f^{\prime }=g$. The displayed estimate proves uniform convergence wherever $|x-c|$ has a common finite bound, and therefore proves locally uniform convergence on any interval.

For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its derivative is $x+1/n$, with supremum error $1/n$ from $g(x)=x$, and $f_n(0)\to 3$. The reconstructed uniform limit is $f(x)=3+x^2/2$; the exact supremum error is $3/n$, equal to the bound above. Without an anchor, the constant functions $f_n(x)=n$ have zero derivatives but no finite pointwise limit. On the real line, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent derivatives, but its supremum error from zero is infinite. The bounded-distance condition has a real role.