Exercises — Uniform convergence of function sequences

Move 1 — pointwise limit. Fix $x\in [0,1]$. As $n\to \infty$, $\frac{x}{n}\to 0$. So $f(x)=0$ for every $x$. (No case-splitting needed — the same happens for all $x$.)

Move 2 — worst gap. The gap is $|f_n(x)-f(x)|=\frac{x}{n}$. On $[0,1]$ this grows with $x$, so it is largest at $x=1$.

Move 3 — sup. $M_n={\sup }_{x\in [0,1]}\frac{x}{n}=\frac{1}{n}\to 0.$ Since $M_n\to 0$, the convergence is uniform on $[0,1]$. $■$

Move 1. Fix any $x\ge 0$: $\frac{x}{n}\to 0$, so still $f(x)=0$ pointwise.

Move 2. The gap $\frac{x}{n}$ has no largest point now — as $x\to \infty$ it grows without bound.

Move 3. $M_n={\sup }_{x\ge 0}\frac{x}{n}=+\infty \text{for every }n.$ $M_n\nrightarrow 0$ (it is never even finite), so convergence is not uniform on $[0,\infty )$.

Moral: same formula, different domain, different answer — uniformity is a property of $(f_n,E)$ together, not of the formula alone. $■$

Move 1 — pointwise. For $x\in [0,1)$, $x^n\to 0$ so $x^n(1-x)\to 0$. At $x=1$, the factor $(1-x)=0$ so $f_n(1)=0$. Hence $f(x)=0$ everywhere.

Move 2 — moving peak. The graph rises then falls; the peak moves toward $1$ as $n$ grows. Differentiate to find it: $f_n^{\prime }(x)=n{x}^{n-1}(1-x)-x^n=x^{n-1}(n-(n+1)x).$ Setting the bracket to $0$ gives the interior maximum at $x_n=\frac{n}{n+1}$.

Move 3 — sup value. $M_n=f_n(x_n)=\left(\frac{n}{n+1}{)}^n\right(1-\frac{n}{n+1})=(\frac{n}{n+1}{)}^n\cdot \frac{1}{n+1}.$ Now $(\frac{n}{n+1}{)}^n=(1+\frac{1}{n}{)}^{-n}\to e^{-1}$, and $\frac{1}{n+1}\to 0$. Therefore $M_n\to e^{-1}\cdot 0=0.$ So the convergence is uniform on $[0,1]$. $■$

The extra factor $(1-x)$ crushes the troublesome peak height to $0$ — contrast with plain $x^n$ where the peak stayed at height $1$.

Move 1 — pointwise. At $x=0$: $f_n(0)=0$. For fixed $x>0$: numerator $\sim nx$, denominator $\sim n^2{x}^2$, so $f_n(x)\sim \frac{1}{nx}\to 0$. Hence $f(x)=0$ everywhere.

Move 2 — moving peak. Substitute $u=nx$: $f_n=\frac{u}{1+u^2}$, which peaks at $u=1$, i.e. $x_n=\frac{1}{n}\in [0,1]$ for all $n\ge 1$.

Move 3 — sup value. At $x_n=1/n$: $M_n=\frac{n\cdot \frac{1}{n}}{1+n^2\cdot \frac{1}{n^2}}=\frac{1}{1+1}=\frac{1}{2}\text{for every }n.$ $M_n=\frac{1}{2}\nrightarrow 0$, so convergence is not uniform on $[0,1]$. A spike of fixed height $\frac{1}{2}$ slides toward $0$ and never shrinks. $■$

Move 1 — pointwise. At $x=0$: $f_n(0)=0$. For fixed $x>0$: as $n\to \infty$, $(1-nx{)}^2\to \infty$, so the denominator blows up and $f_n(x)\to 0$. Hence $f(x)=0$.

Move 2 — clever bad point. The denominator is smallest when $1-nx=0$, i.e. $x=\frac{1}{n}$. Choose $x_n=\frac{1}{n}$ (which lies in $[0,1]$).

Move 3 — evaluate. $f_n\left(\frac{1}{n}\right)=\frac{(1/n{)}^2}{(1/n{)}^2+0}=1.$ So $M_n\ge f_n(1/n)=1$ for all $n$, hence $M_n\nrightarrow 0$: not uniform.

Notice pointwise convergence holds at each fixed $x$ (freeze $x$, the point $1/n$ eventually passes it by), yet at the moving point $x_n=1/n$ the value is pinned at $1$. This is exactly why uniformity is strictly stronger. $■$

Each $f_n(x)=x^n$ is continuous on $[0,1]$. The pointwise limit is $f(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array},$ which is discontinuous at $x=1$.

The theorem (contrapositive): if a sequence of continuous functions converges uniformly, the limit is continuous. The limit here is not continuous, therefore the convergence cannot be uniform. $■$

This is a "detect the crime by its footprint" argument (see Continuity preserved under uniform limits): a discontinuous limit of continuous functions is proof of non-uniformity, no sup computation required.

Move 1 — pointwise limit on $[0,\frac{1}{2}]$. Every $x\in [0,\frac{1}{2}]$ has $x<1$, so $x^n\to 0$; thus $f\equiv 0$ here.

Move 2 — uniformity. $M_n={\sup }_{x\in [0,1/2]}{x}^n={\left(\frac{1}{2}\right)}^n\to 0$, so convergence is uniform. By the Interchange of limit and integral theorem, we may swap $\lim$ and $\int$.

Move 3 — check both sides explicitly.

• Right side: ${\int }_0^{1/2}0dx=0.$
• Left side: ${\int }_0^{1/2}{x}^{n}dx=\frac{(1/2{)}^{n+1}}{n+1}$, and this $\to 0$ as $n\to \infty$.

Both equal $0$. The interchange is valid, exactly as the theorem promised. $■$

Move 1 — pointwise limit. At $x=0$ and $x=1$, $g_n=0$. For fixed $x\in (0,1)$: $(1-x{)}^n\to 0$ geometrically, which beats the polynomial $n^2$, so $g_n(x)\to 0$. Hence $g\equiv 0$ pointwise.

Move 2 — the integral. Use ${\int }_0^{1}x(1-x{)}^{n}dx=\frac{1}{(n+1)(n+2)}$ (Beta integral). Therefore ${\int }_0^1{g}_{n}dx=n^2\cdot \frac{1}{(n+1)(n+2)}=\frac{n^2}{(n+1)(n+2)}\to 1.$

Move 3 — interpret. The pointwise-limit integral is $0$, but ${\lim }_n\int g_n=1\ne 0$. The interchange fails. By the contrapositive of the Interchange Theorem, $g_n$ does not converge uniformly on $[0,1]$. Indeed a tall thin spike (height $\sim n^2\cdot \frac{1}{n}=n$ near $x\sim \frac{1}{n}$) carries a fixed amount of area to the finish line. $■$

Contrast with the Dominated convergence theorem: even that softer tool would fail here because there is no integrable function dominating all $g_n$.

Move 1 — bound the tail uniformly. For $m>n$, $|f_m(x)-f_n(x)|=\left|{\sum }_{k=n+1}^m\frac{\cos (kx)}{k^2}\right|\le {\sum }_{k=n+1}^m\frac{|\cos (kx)|}{k^2}\le {\sum }_{k=n+1}^m\frac{1}{k^2}.$ The bound $\frac{1}{k^2}$ is independent of $x$ — this is the Weierstrass $M$-test in action, with $M_k=\frac{1}{k^2}$ and $\sum M_k=\frac{{\pi }^2}{6}<\infty$.

Move 2 — take the sup, then send $n\to \infty$. Since the bound has no $x$, ${\sup }_{x\in \mathbb{R}}|f_m(x)-f_n(x)|\le {\sum }_{k=n+1}^{\infty }\frac{1}{k^2}=:R_n.$ Because $\sum 1/k^2$ converges, its tail $R_n\to 0$. So given $\epsilon >0$ we pick $N$ with $R_N<\epsilon$; then for all $m,n\ge N$ the sup-distance is $<\epsilon$.

Move 3 — conclude. That is exactly the uniform Cauchy criterion, so $(f_n)$ converges uniformly on $\mathbb{R}$ — even though we never wrote the limit function down. (See Cauchy sequences in metric spaces: $C(\mathbb{R})$ with sup-norm is complete, so uniform-Cauchy $\Rightarrow$ uniform-convergent.) $■$

Move 1 — pointwise. At $x=0$, $f_n(0)=0$. For fixed $x>0$, the exponential $e^{-n{x}^2}\to 0$ beats the linear factor $nx$, so $f_n(x)\to 0$. Hence $f\equiv 0$.

Move 2 — moving peak. Differentiate in $x$: $f_n^{\prime }(x)=n{e}^{-n{x}^2}(1-2n{x}^2)=0⟹x_n=\frac{1}{\sqrt{2n}}.$ Peak height: $M_n=f_n(x_n)=n\cdot \frac{1}{\sqrt{2n}}{e}^{-1/2}=\sqrt{\frac{n}{2}}{e}^{-1/2}\to \infty .$ $M_n\to \infty \nrightarrow 0$: convergence is not uniform (a spike that grows and narrows).

Move 3 — the integrals. Substitute $u=n{x}^2$, $du=2nxdx$: ${\int }_0^{\infty }nx{e}^{-n{x}^2}dx=\frac{1}{2}{\int }_0^{\infty }{e}^{-u}du=\frac{1}{2}\text{for every }n.$ So $\underset{n}{\lim }{\int }_0^{\infty }{f}_{n}dx=\frac{1}{2}$, but ${\int }_0^{\infty }\underset{n}{\lim }{f}_{n}dx={\int }_0^{\infty }0dx=0$.

Since $\frac{1}{2}\ne 0$, the interchange fails — consistent with the non-uniformity found in Move 2. A fixed area $\frac{1}{2}$ rides the shrinking spike all the way out. $■$

• $x/n$ on $[0,1]$ vs $[0,\infty )$? ::: $M_n=1/n\to 0$ (uniform) vs $M_n=\infty$ (not uniform).
• $x^n(1-x)$ on $[0,1]$? ::: $M_n=(n/(n+1){)}^n/(n+1)\to 0$, uniform.
• $nx/(1+n^2{x}^2)$ on $[0,1]$? ::: Peak at $x_n=1/n$, $M_n=1/2$, not uniform.
• Fastest non-uniformity proof for $x^n$ on $[0,1]$? ::: Limit is discontinuous at $1$; contrapositive of continuity theorem.
• $nx{e}^{-n{x}^2}$ on $[0,\infty )$: is $\lim \int =\int \lim$? ::: No: $1/2\ne 0$; spike carries fixed area.
• Tool to prove uniformity without the limit? ::: Uniform Cauchy criterion / Weierstrass $M$-test.