Question bank — Uniform convergence of function sequences

If $f_n\to f$ uniformly on $E$, then $f_n\to f$ pointwise on $E$. ::: True. Uniform gives one $N$ that works for all $x$, so in particular it works at each fixed $x$ — that is exactly pointwise. Uniform is strictly stronger, never weaker. If $f_n\to f$ pointwise and each $f_n$ is continuous, then $f$ is continuous. ::: False. $x^n$ on $[0,1]$ converges pointwise to a function that jumps at $x=1$. You need uniform convergence for the $\epsilon /3$ argument to work. $f_n(x)=x^n$ converges uniformly on $[0,a]$ for every fixed $a<1$. ::: True. There the sup is attained at the right endpoint, $M_n=a^n\to 0$. The trouble point $x=1$ is excluded, so uniformity is restored. Uniform convergence on $[0,a]$ for every $a<1$ implies uniform convergence on $[0,1)$. ::: False. "Uniform on each piece" is not "uniform on the union". As $a\to 1$ the required $N$ blows up; on all of $[0,1)$ the sup of $x^n$ is $1$ for every $n$. If $M_n\to 0$ then $f_n\to f$ uniformly, regardless of which $f$ you used. ::: False as stated — you must use the true pointwise limit $f$. Any other target function makes $M_n$ the wrong quantity, and it may not even go to $0$. Uniform convergence of $f_n$ implies uniform convergence of $f_n^{\prime }$. ::: False. $\frac{\sin (nx)}{n}\to 0$ uniformly, yet $f_n^{\prime }=\cos (nx)$ does not converge at all. Differentiation amplifies fast wiggles. If $f_n\to f$ and $g_n\to g$ both uniformly on $E$, then $f_n+g_n\to f+g$ uniformly. ::: True. $\sup |(f_n+g_n)-(f+g)|\le \sup |f_n-f|+\sup |g_n-g|\to 0$ by the triangle inequality on the sup. If $f_n\to f$ and $g_n\to g$ uniformly on $E$, then $f_n{g}_n\to fg$ uniformly on $E$. ::: False in general. Products need boundedness: on unbounded $E$, $f_n(x)=x+\frac{1}{n}$ times $g_n=x$ shows the error $\frac{x}{n}$ has infinite sup. It is true when the limits are bounded. The uniform Cauchy criterion lets you prove uniform convergence without knowing the limit. ::: True. If ${\sup }_x|f_n-f_m|\to 0$ as $m,n\to \infty$, the sequence converges uniformly to some $f$; completeness of the sup-norm space supplies the limit. Uniform convergence preserves boundedness of the functions. ::: True (for large $n$). If $|f_n-f|<1$ for all $x$ once $n\ge N$ and $f$ is bounded, then each such $f_n$ is bounded; and $f$ itself is bounded whenever the $f_n$ are uniformly bounded.

"$f_n(x)=x^n$ converges to $0$ on $[0,1]$, so ${\int }_0^1{f}_n\to {\int }_0^{1}0=0$." ::: The integral conclusion is numerically right (${\int }_0^1{x}^n=\frac{1}{n+1}\to 0$), but the justification is wrong: convergence is not uniform, so you cannot invoke the interchange theorem. The value happens to agree for another reason. "Convergence is uniform because for each $x$ I found an $N(x)$ that works." ::: Finding $N$ per point is only pointwise. Uniform needs a single $N$ independent of $x$ — you must show ${\sup }_{x}N(x)$ is finite, i.e. bound the error uniformly. "$M_n={\sup }_{x\in [0,1)}{x}^n=x^n$ evaluated at the largest $x$, but $[0,1)$ has no largest point, so $M_n$ is undefined." ::: The sup need not be attained. ${\sup }_{x\in [0,1)}{x}^n=1$ as a supremum (approached as $x\to 1^{-}$) even though no $x$ hits it. Sup ≠ max. "Since $|f_n(x)-f(x)|<\epsilon$ for all $x$, we have $M_n<\epsilon$, and equality can never happen." ::: You may only conclude $M_n\le \epsilon$; the strict inequality can degrade to $\le$ when you take a sup. That is why the definition-to-sup proof uses $\epsilon /2$, so $M_n\le \epsilon /2<\epsilon$. "The $\epsilon /3$ proof of continuity only used continuity of the $f_n$, not uniformity." ::: It secretly used uniformity in step 1: the same $N$ must control $|f-f_N|$ at both $x$ and $x_0$ at once. Pointwise would give two different $N$'s and the argument collapses. "$f_n\to f$ pointwise with $f$ continuous, therefore convergence is uniform." ::: A continuous limit is necessary but not sufficient. $f_n(x)=\frac{nx}{1+n^2{x}^2}$ on $[0,1]$ converges pointwise to the continuous function $0$, yet the peak stays at height $\frac{1}{2}$, so $M_n\nrightarrow 0$. "Uniform convergence on $\mathbb{R}$ means the graphs eventually coincide." ::: They stay within a shrinking horizontal band of width $2\epsilon$, not identical. "Coincide" would mean $f_n=f$ exactly; uniform only says the maximum vertical gap shrinks to $0$.

Why does swapping the quantifiers to $\forall \epsilon \exists N\forall x$ make uniform strictly harder than pointwise? ::: Placing $\exists N$ before $\forall x$ forces a single $N$ to serve every point simultaneously, so it must handle the worst point too. Pointwise lets slow points pick their own large $N$. Why is "$\sup$ of the gap $\to 0$" the natural test rather than checking each point? ::: The phrase "for all $x$, gap $<\epsilon$" is the statement that the largest gap is $<\epsilon$, and the largest gap is precisely the supremum. So uniform convergence collapses infinitely many point conditions into one number, $M_n$. Why does uniform convergence let you interchange limit and integral, but pointwise generally does not? ::: The single bound $M_n$ controls the whole integrand, giving $|\int f_n-\int f|\le (b-a)M_n\to 0$. Pointwise offers no uniform bound, so mass can escape (a spike of fixed area can slide off). Why does the counterexample $\sin (nx)/n$ show derivatives are not protected? ::: The functions shrink because of the $\frac{1}{n}$ factor, but differentiating cancels it: $\cos (nx)$ has amplitude $1$ forever. Uniform closeness of heights says nothing about slopes. Why is uniform Cauchy convergence useful for series like in the Weierstrass M-test? ::: You can bound ${\sup }_x|f_n-f_m|$ using tail sums of a numeric series before identifying the limit function, then completeness delivers the limit. This is exactly how the M-test certifies uniform convergence. Why does uniformity depend on the domain, not just the formula? ::: The sup is taken over the domain; enlarging the domain can expose a troublesome point (like $x\to 1$ for $x^n$) that inflates $M_n$. Same formula, different sup, different verdict. Why must you compute the pointwise limit first before testing uniformity? ::: $M_n$ is defined as $\sup |f_n-f|$, so you literally need $f$ to write it down. Guessing a wrong $f$ measures the wrong gap and gives a meaningless answer.

Does uniform convergence on the empty set or a single point make sense? ::: Yes, trivially. On a one-point set pointwise and uniform coincide (one point can't disagree with itself), and on the empty set the sup over an empty set is $0$, so every sequence converges uniformly. What happens with $f_n(x)=x^n$ exactly at the boundary point $x=1$? ::: $1^n=1$ for all $n$, so the sequence is constant $1$ there and converges to $1$, not $0$. This lone point is what makes the pointwise limit discontinuous and destroys uniformity on $[0,1]$. Is a constant sequence $f_n=g$ (same function for all $n$) uniformly convergent? ::: Yes. $M_n=\sup |g-g|=0$ for every $n$, so it converges uniformly to $g$ trivially — the strongest possible case. If each $f_n$ is discontinuous but $f_n\to f$ uniformly, must $f$ be discontinuous? ::: No. The continuity theorem is one-directional: continuous $f_n$ force continuous $f$, but discontinuous $f_n$ can have a continuous (or discontinuous) uniform limit; uniformity says nothing forcing a break. On an unbounded domain, can a sequence be uniform even though each $f_n$ is unbounded? ::: Yes. $f_n(x)=x+\frac{1}{n}$ on $\mathbb{R}$ has each $f_n$ unbounded, yet $M_n={\sup }_x|f_n-x|=\frac{1}{n}\to 0$, so it converges uniformly to $f(x)=x$. What is $M_n$ when the pointwise limit does not exist at some point? ::: Then $f$ isn't defined there, so uniform convergence fails at the definition stage — there is no target to be uniformly close to. Uniform convergence presupposes pointwise convergence everywhere. Degenerate case: if $M_n\to 0$ but is never exactly $0$, is that still uniform? ::: Yes. Uniform convergence only requires the limit of $M_n$ to be $0$; it never demands the functions actually equal $f$. A strictly positive sequence tending to $0$ is perfectly uniform.