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Composition with $t\mapsto t^{1+{\alpha }_n}$ preserves continuity, so $A_n$ maps the space of continuous functions on a compact space $C([0,1])$ into itself and is linear. Since $t\mapsto t^{1+{\alpha }_n}$ maps $[0,1]$ onto itself,For $0<t\le 1$, the mean value theorem applied to $r\mapsto t^{1+r}$ givesThe same bound holds at $t=0$. Each $x\in C([0,1])$ has uniform continuity by the Heine-Cantor theorem. With its modulus of continuity ${\omega }_x(\delta )={\sup }_{|s-t|\le \delta }|x(s)-x(t)|$, we obtainAlternatively, the bound proves convergence on every polynomial; the Weierstrass approximation theorem makes these a dense subset of $C([0,1])$, and the Banach-Steinhaus convergence theorem with $\Vert A_n\Vert =1$ gives the same conclusion.

Use the modulus of continuity on the circle, $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. Splitting a displacement into steps of length at most $\delta$ and applying the triangle inequality givesThe Fejér kernel is nonnegative and has integral one, soWe therefore need its first absolute moment. The inequalities $|\sin (nt/2)|\le n|\sin (t/2)|$ and $\sin (|t|/2)\ge |t|/\pi$ on $[-\pi ,\pi ]$ implySplitting at $1/n$ givesTaking $\delta ={\delta }_n=(\ln n)/n$ in the preceding error bound proves the Fejér first-moment approximation bound:The constants do not depend on $f$ or $n$. By the Heine-Cantor theorem, a continuous function on the compact circle is uniformly continuous, hence $\omega (f,{\delta }_n)\to 0$ as ${\delta }_n\to 0$. Thus the Fejér sums converge uniformly to $f$. The index restriction is necessary: at $n=1$ the printed scale is zero, while ${\sigma }_{1}f$ is the mean of $f$ and generally differs from it.

Let $L$ be the supremum of all lower Darboux sums and $U$ the infimum of all upper Darboux sums. The assumed comparison of arbitrary lower and upper sums gives $L\le U$. For the partition supplied for a given $\epsilon$,Since this holds for every positive $\epsilon$, $U=L$. Thus the Riemann integrability criterion gives

If $f$ is continuous on $[a,b]$, the Heine-Cantor theorem makes it uniformly continuous. Given $\epsilon >0$, choose $\delta$ so that intervals of length below $\delta$ have oscillation below $\epsilon /(b-a)$. A partition of mesh below $\delta$ then satisfiesproving that Continuous functions are Riemann integrable.

For the function defined using $g$, choose $K$ with $|f|\le K$. Given $\epsilon >0$, choose $\delta >0$ so small that $2K\delta <\epsilon /2$. On $[\delta ,1]$, the function $g$ is continuous and therefore has a partition whose upper-minus-lower sum is below $\epsilon /2$. Adding the interval $[0,\delta ]$ contributes at most $2K\delta$, regardless of the value $f(0)=\lambda$. Hence the full Darboux-sum difference is below $\epsilon$, and

Finally, for any partition $a=x_0<\cdots <x_n=b$, the mean value theorem supplies ${\xi }_i\in (x_{i-1},x_i)$ such thatSumming telescopes toa Riemann sum for the Riemann-integrable function $f^{\prime }$. Equivalently, every lower derivative sum is at most this telescoping value and every upper derivative sum is at least it. Taking the common upper and lower Darboux integrals proves the fundamental theorem of calculus conclusion

True. Define $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one. The restriction of $f$ to $[-n,n]$ is uniformly continuous by the Heine-Cantor theorem; its modulus of continuity combined with $|c_n(x)-c_n(y)|\le |x-y|$ proves uniform continuity of $f_n$ on all of $\mathbb{R}$. For each fixed $x$, once $n\ge |x|$ we have $f_n(x)=f(x)$. HenceThis pointwise approximation by uniformly continuous functions even gives eventual equality on every fixed bounded interval. It need not give uniform convergence on the whole line, as the bounded counterexample in part (i) already demonstrates.

A real function on a metric space is uniformly continuous if for every $\epsilon >0$ there is a $\delta >0$ such that, simultaneously for all $x,y$, $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$. For a continuous function on a finite closed interval $[a,b]$, suppose this fails. Then there are ${\epsilon }_0>0$ and pairs $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. By compactness, some subsequence $x_{n_j}$ converges to $x\in [a,b]$; then $y_{n_j}\to x$ too. Continuity contradicts the image separation. This proves the Heine-Cantor theorem in this setting. Boundedness of the interval matters: $x^2$ on $[0,\infty )$ is a counterexample if “closed interval” includes unbounded intervals.

A function is Lipschitz continuous if some finite $K\ge 0$ satisfies $|f(x)-f(y)|\le Kd(x,y)$ for every pair. Choosing $\delta =\epsilon /(K+1)$ proves uniform continuity.

The first assertion is false. Take $f(x)=\sin (x^2)$, which is bounded and continuous, andThen $f(x_n)=1$, $f(y_n)=-1$, while $y_n-x_n=\pi /(y_n+x_n)\to 0$. This violates uniform continuity.

This is always true: it is the Heine-Cantor theorem. If uniform continuity failed, there would be an $\epsilon >0$ and sequences $x_n,y_n\in X$ such thatBy sequential compactness of a compact metric space, some subsequence $x_{n_k}$ converges to $x\in X$. The triangle inequality gives $y_{n_k}\to x$. Continuity then makes both image subsequences converge to $f(x)$, contradicting their separation by $\epsilon$.

Suppose the continuous function $f$ did not have uniform continuity on $[a,b]$. Negating its definition gives ${\epsilon }_0>0$ such that for every $n\ge 1$ there are $x_n,y_n\in [a,b]$ withThe Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence, gives $x_{n_k}\to p$. The inequalities $a\le x_{n_k}\le b$ imply $p\in [a,b]$. By the triangle inequality,Since continuous functions preserve convergent sequences, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its absolute value being at least ${\epsilon }_0$. This proves the Heine-Cantor theorem in the required setting: $f$ has uniform continuity on $[a,b]$.

By the Heine-Cantor theorem, a continuous function on the compact interval $[a,b]$ is uniformly continuous. Given $\epsilon >0$, choose $\delta >0$ such that $|f(s)-f(t)|<\epsilon$ whenever $s,t\in [a,b]$ and $|s-t|<\delta$.

Choose a finite partition of an interval $a=t_0<t_1<\cdots <t_m=b$ with every width below $\delta$. Define the step functionFor each $t<b$, its distance from the chosen left endpoint is below $\delta$, so $|f(t)-\varphi (t)|<\epsilon$; at $b$ the error is zero. HenceA step function may have separately assigned values at its finitely many partition points; these values do not affect its Riemann integral. If $a=b$, simply take $\varphi =f$ on that singleton.

Every continuous $f:\mathbb{R}\to \mathbb{R}$ can be approximated with pointwise convergence by uniformly continuous functions. Set $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one and $f{|}_{[-n,n]}$ is uniformly continuous by the Heine-Cantor theorem, so $f_n$ is uniformly continuous on the whole line. On every fixed bounded interval, $f_n=f$ eventually.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.

A continuous function on a compact interval is uniformly continuous by the Heine-Cantor theorem. Choose a partition of an interval whose widths lie below the input tolerance for a desired output error $\epsilon$. On each half-open partition subinterval, use the value of the function at its left endpoint, and assign the rightmost endpoint its own value. This defines a step function whose error is strictly below $\epsilon$ at every input. Finitely assigned partition-point values are allowed and do not alter the Riemann integral.