Sequences and Series of Functions; Uniform Convergence: Examples and Counterexamples

If $0\le x<1$, then $|x|<1$, so [L1] gives $x^{k+1}\to0=\chi(x)$; at $x=1$, one has $f_k(1)=1=\chi(1)$ for every $k$. Thus $f_k\to\chi$ pointwise.

For each $k$, put $a_k:=\iota(k+1)>0$ and $y_k:=1-1/(2a_k)$. Then $0\le y_k<1$, so $\chi(y_k)=0$.

Apply [L2] with $n=k+1$ and $t=-1/(2a_k)$: $f_k(y_k)=\left(1-1/(2a_k)\right)^{k+1}\ge1-a_k/(2a_k)=1/2$.

Hence $|f_k(y_k)-\chi(y_k)|\ge1/2$ for every $k$, so no index makes the error smaller than $1/2$ at every point; the convergence is not uniform.

The functions therefore satisfy the refuted claim's hypothesis and violate its conclusion.

Each $r_k$ is bounded and is continuous except possibly at $0$ and $1/a_k$, so it is integrable by [L2].

Let $q_k$ equal $a_k$ on $[0,1/a_k]$ and $0$ on $(1/a_k,1]$. The functions $q_k$ and $r_k$ differ only at $0$, so they have the same integral by [L3].

At $x=0$ one has $r_k(0)=0$ for all $k$. If $x>0$, choose $N$ with $1/\iota(N)<x$; for $k\ge N$, monotonicity of the canonical naturals gives $1/a_k<x$, hence $r_k(x)=0$. Thus $r_k\to0$ pointwise.

To see explicitly that the convergence is not uniform, take $\varepsilon:=1/2$. For every proposed $N\in\mathbb N$, choose $k:=N$ and $x_N:=1/a_N$; then $|r_N(x_N)-0|=a_N\ge1>\varepsilon$. Thus the uniform quantifier condition in [L5] fails.

By [L3] and [L4], endpoint values do not affect either piece, and splitting at $1/a_k$ when it lies in the interior, with the coincident-endpoint convention otherwise, gives $\int_0^1q_k=a_k(1/a_k)+0=1$.

Steps 1.2 and 2.1 give $\int_0^1r_k=1$ for every $k$, whereas the integral of the zero function is $0$.

The sequence therefore converges pointwise to $0$ but its integrals do not converge to the integral of the limit, refuting the claim.

For every $k$, the denominator $a_k+x$ is positive on $[0,\infty)$, so $f_k$ is continuous by [L1]; the zero function is continuous as well.

Since $a_{k+1}>a_k>0$, one has $a_{k+1}+x>a_k+x>0$, hence $f_{k+1}(x)\le f_k(x)$ for every $x\ge0$.

Fix $x\ge0$. If $x=0$ then $f_k(x)=0$; if $x>0$, then $0\le f_k(x)\le x/a_k$, and [L2] gives $x/a_k\to0$. Thus $f_k(x)\to0$ for every $x$.

At $x_k:=a_k$ one has $f_k(x_k)=a_k/(a_k+a_k)=1/2$, so the convergence is not uniform.

The domain $[0,\infty)$ is not compact by [L3].

Hence all the listed Dini hypotheses except compactness hold, while the uniform conclusion fails; compactness cannot be dropped.

The intervals $(0,1/a_{k+1})$ are contained in $(0,1/a_k)$, so $h_{k+1}(x)\le h_k(x)$ for every $x\in[0,1]$.

At $x=0$, every $h_k(x)$ is $0$. If $x>0$, choose $N$ with $1/\iota(N)<x$; then $h_k(x)=0$ for all $k\ge N$. Thus $h_k\to0$ pointwise.

Each $h_k$ is discontinuous at $0$: for any $\delta>0$, the point $y:=\min\{\delta/2,1/(2a_k)\}$ satisfies $0<y<\delta$, lies in $(0,1/a_k)$, and has $|h_k(y)-h_k(0)|=1$.

For each $k$, the point $x_k:=1/(2a_k)$ lies in $(0,1/a_k)$ and satisfies $h_k(x_k)=1$, so the convergence to $0$ is not uniform.

The compact domain, monotone pointwise convergence, and continuous limit remain, but the approximants are discontinuous and uniform convergence fails; their continuity is indispensable in [L3].

Each $f_k$ is continuous by [L2].

For $x\in[0,1]$, $f_{k+1}(x)=x^{k+2}=x\,x^{k+1}\le x^{k+1}=f_k(x)$, so the sequence is pointwise nonincreasing.

The pointwise convergence to $\chi$ and the failure of uniform convergence are [L1].

The function $\chi$ is discontinuous at $1$: for every $\delta>0$, the point $y:=1-\min\{\delta/2,1/2\}$ lies in $[0,1)$ with $|y-1|<\delta$ and $|\chi(y)-\chi(1)|=1$.

Thus compactness, continuity of all approximants, and monotonicity hold, but the limit is discontinuous and the uniform conclusion fails; continuity of the limit in [L3] is indispensable.

Every $h_k$ is continuous by [L1], and the zero function is continuous.

If $x=0$, then $h_k(x)=0$. If $x>0$, choose $N$ with $1/\iota(N)<x$; for all sufficiently large $k$, $a_kx\ge1$, so $|2a_kx-1|\ge1$ and $h_k(x)=0$. Thus $h_k\to0$ pointwise.

At $x_k:=1/(2a_k)$ one has $h_k(x_k)=1$, so the convergence is not uniform.

At $x=1/4$, the values at $k=0,1,2$ are respectively $1/2,1,1/2$, so the sequence is neither pointwise nondecreasing nor pointwise nonincreasing.

All Dini hypotheses except monotonicity hold while uniform convergence fails, so monotonicity cannot be dropped.

The denominator $\iota(m+1)+\iota(n+1)$ is positive for all $m,n$, so every $s_{m,n}$ is defined and lies between $0$ and $1$.

Fix $m$ and put $a:=\iota(m+1)>0$. Then $0\le s_{m,n}\le a/\iota(n+1)$; given $\varepsilon>0$, [L2] makes the latter smaller than $\varepsilon$ for all sufficiently large $n$. Hence $\lim_ns_{m,n}=0$.

Fix $n$ and put $b:=\iota(n+1)>0$. Since $|1-s_{m,n}|=b/(\iota(m+1)+b)\le b/\iota(m+1)$, [L2] makes this smaller than any prescribed $\varepsilon>0$ for all sufficiently large $m$. Hence $\lim_ms_{m,n}=1$.

By step 1.2 the first inner-limit sequence is constantly $0$, so its limit in $m$ is $0$; by step 1.3 the other inner-limit sequence is constantly $1$, so its limit in $n$ is $1$.

Thus both iterated limits exist and are unequal, refuting the claim.

The denominator $1+a_k^2x^2$ is positive for every $x$, so $f_k$ is differentiable on $\mathbb{R}$ by [L2].

From $(a_k|x|-1)^2\ge0$ one obtains $2a_k|x|\le1+a_k^2x^2$, hence $|f_k(x)|\le1/(2a_k)$ for every $x\in\mathbb{R}$.

At $x=0$, the numerator $x$ has derivative $1$, the denominator $1+a_k^2x^2$ has value $1$ and derivative $0$, so the quotient rule gives $f_k'(0)=1$.

The zero function has derivative $0$ by the constant case of the power rule.

Given $\varepsilon>0$, [L3] gives $N$ such that $1/(2a_k)<\varepsilon$ for every $k\ge N$; step 1.2 then gives $|f_k(x)|<\varepsilon$ for every $x\in\mathbb{R}$.

Step 2.1 proves $f_k\to0$ uniformly, while steps 1.3 and 1.4 show that the derivatives at $0$ do not converge to the derivative of the limit.

The uniformly convergent differentiable sequence therefore refutes the claim.