For each \(x\) in \([0,1]\) the numbers \(x^n\) converge as \(n\to\infty\) : to \(0\) if \(x<1\) , and to \(1\) if \(x=1\) . Each function \(x^n\) is continuous, indeed a polynomial. Their limit is not continuous. In symbols,

The two limits do not commute. Cauchy stated in 1821 that a convergent series of continuous functions has a continuous sum, and the statement stood for years before the hidden assumption was identified.

The same difficulty appears three times in analysis, since continuity, the integral, and the derivative are all defined by limits. Is the limit of continuous functions continuous? Is the integral of the limit the limit of the integrals? Is the derivative of the limit the limit of the derivatives? For convergence at each point separately, the answer to all three is no. This chapter isolates the stronger notion of convergence under which the first two answers become yes and the third becomes yes with an extra hypothesis, and then applies it to series of functions, where it pays for itself at once: power series can be differentiated and integrated term by term.

8.1 Pointwise convergence and its failures

Here \(N\) is allowed to depend on \(x\) as well as on \(\varepsilon\) . That freedom is the source of every failure below.

8.2 Uniform convergence

Now \(N\) depends on \(\varepsilon\) only. Geometrically: for \(n\ge N\) the whole graph of \(f_n\) lies in the band of half-width \(\varepsilon\) around the graph of \(f\) . Uniform convergence implies pointwise convergence to the same limit, so the only candidate for a uniform limit is the pointwise one. The relation between the two notions is that between continuity and uniform continuity in Chapter 5: the same words with one quantifier moved.

For a function \(g\) on \(S\) write \(\lVert g\rVert_\infty=\sup_{x\in S}\lvert g(x)\rvert\) , which may be \(+\infty\) . For bounded functions this is the distance from \(g\) to \(0\) in the metric space \(B(S)\) of Proposition 4.4.

This turns a statement with three quantifiers into the convergence of one sequence of numbers. To test for uniform convergence: find the pointwise limit \(f\) , compute or estimate \(M_n=\lVert f_n-f\rVert_\infty\) , and see whether \(M_n\to 0\) .

Example 8.2: Computing the supremum

Powers on a shorter interval. On \([0,a]\) with \(0<a<1\) , \(\lVert x^n-0\rVert_\infty=a^n\to 0\) . So \(x^n\to 0\) uniformly on \([0,a]\) . On \([0,1)\) , however, \(\sup_{0\le x<1}x^n=1\) for every \(n\) , so the convergence to \(0\) is not uniform there, even though the limit function \(0\) is continuous on \([0,1)\) . Uniformity depends on the set.

A sequence on the whole line. Let \(f_n(x)=x/(1+nx^2)\) on \(\mathbb R\) . The pointwise limit is \(0\) . From \(1+nx^2\ge 2\sqrt n\,\lvert x\rvert\) , which is \((1-\sqrt n\lvert x\rvert)^2\ge 0\) , we get

\[ \lvert f_n(x)\rvert\le\frac{1}{2\sqrt n}\qquad\text{for all }x, \]

with equality at \(x=1/\sqrt n\) . So \(\lVert f_n\rVert_\infty=1/(2\sqrt n)\to 0\) , and \(f_n\to 0\) uniformly on \(\mathbb R\) . For \(n=100\) the largest value anywhere is \(0.05\) .

Theorem 8.4 (uniform Cauchy criterion). A sequence \((f_n)\) of real functions on \(S\) converges uniformly on \(S\) if and only if for every \(\varepsilon>0\) there is an \(N\) such that \(\lvert f_n(x)-f_m(x)\rvert<\varepsilon\) for all \(m,n\ge N\) and all \(x\in S\) .

Proof. If \(f_n\to f\) uniformly, choose \(N\) with \(\lvert f_n(x)-f(x)\rvert<\varepsilon/2\) for all \(n\ge N\) and all \(x\) , and use the triangle inequality. Conversely, suppose the condition holds. For each fixed \(x\) the sequence \((f_n(x))\) is a Cauchy sequence of real numbers, so it converges; call its limit \(f(x)\) . Given \(\varepsilon>0\) take \(N\) as in the condition for \(\varepsilon/2\) . For \(n\ge N\) and any \(x\) , let \(m\to\infty\) in \(\lvert f_n(x)-f_m(x)\rvert<\varepsilon/2\) to get \(\lvert f_n(x)-f(x)\rvert\le\varepsilon/2<\varepsilon\) . The same \(N\) works for every \(x\) , so the convergence is uniform.

Restricted to bounded functions, the theorem says that \(B(S)\) with the supremum metric is a complete metric space: the pointwise completeness of \(\mathbb R\) has been lifted to a space of functions.

Common mistake. Proving pointwise convergence and calling it uniform. An argument that begins “fix \(x\) ; given \(\varepsilon\) choose \(N\) such that…” has let \(N\) depend on \(x\) , and proves pointwise convergence only, however the last line is worded. To prove uniform convergence, produce a bound for \(\lvert f_n(x)-f(x)\rvert\) that does not contain \(x\) and tends to \(0\) . To disprove it, show that \(\lVert f_n-f\rVert_\infty\) does not tend to \(0\) , for instance by evaluating at points \(x_n\) that move with \(n\) .

8.3 Continuity of the limit

The first and third terms are controlled at two different points, \(x\) and \(a\) , by the same \(N\) . That is exactly what uniformity provides and what pointwise convergence does not: with pointwise convergence the \(N\) for the point \(x\) would change as \(x\) varies, and no single \(f_N\) could be used.

The theorem is most often used in the contrapositive: since the limit of \(x^n\) on \([0,1]\) is discontinuous, the convergence is not uniform. The converse is false. A sequence of continuous functions can converge non-uniformly to a continuous limit, as the humps \(g_n\) of Example 8.1 and the powers on \([0,1)\) show.

For a compact metric space \(K\) , let \(C(K)\) be the set of continuous real functions on \(K\) . They are bounded, by the Extreme Value Theorem, so \(C(K)\subseteq B(K)\) , and \(C(K)\) is a metric space with the supremum metric. Theorem 8.5 says that \(C(K)\) is a closed subset of \(B(K)\) , and a closed subset of a complete space is complete, since a Cauchy sequence in it converges in the larger space and the limit stays in the subset. So \(C[a,b]\) is a complete metric space. It is the first example of the function spaces studied in Functional Analysis, and such spaces differ from \(\mathbb R^n\) in a way that Example 4.5 has already shown for the space of bounded sequences: their closed bounded sets need not be compact. Exercise 18 shows the same for \(C[0,1]\) itself.

There is one situation in which pointwise convergence to a continuous limit is automatically uniform.

The theorem also holds for increasing sequences, by applying it to \(-f_n\) . Each hypothesis is needed. The powers \(x^n\) on \([0,1]\) are monotone on a compact set, but the limit is not continuous. On \([0,1)\) they are monotone with a continuous limit, but the set is not compact. The humps \(g_n\) have a continuous limit on a compact set but are not monotone in \(n\) .

8.4 Integrals of limits

The last line is worth remembering on its own: the integral is a Lipschitz function on \(C[a,b]\) with respect to the supremum metric, with constant \(b-a\) .

8.5 Derivatives of limits

For derivatives uniform convergence is not enough, and the failure is more radical than for integrals.

8.6 Series of functions and power series

A series of functions \(\sum f_n\) converges pointwise, or uniformly, on \(S\) if its sequence of partial sums does. Theorems 8.5, 8.7, and 8.8 then say: a uniformly convergent series of continuous functions has a continuous sum; it may be integrated term by term over \([a,b]\) ; and a series of continuously differentiable functions may be differentiated term by term if the differentiated series converges uniformly and the original series converges at one point. The standard test for uniform convergence of a series is a comparison with a series of numbers.

8.7 Approximation by polynomials

Power series approximate only very special functions. Yet every continuous function on a closed bounded interval can be approximated uniformly by polynomials. The polynomials are not Taylor polynomials, which need derivatives, but averages of sampled values.

For a function \(f\) on \([0,1]\) , the Bernstein polynomial of degree \(n\) is

The proof uses three earlier chapters at once: uniform continuity from compactness, a supremum estimate that is uniform in \(x\) , and the idea, from the proof of Theorem 7.8, of splitting a sum into a part where the function varies little and a part that carries little weight.

Exercises

1. Find the pointwise limit and decide whether the convergence is uniform: (a) \(f_n(x)=x/n\) on \([0,1]\) and on \(\mathbb R\) ; (b) \(f_n(x)=nx/(1+nx)\) on \([0,1]\) and on \([1,2]\) ; (c) \(f_n(x)=x^n(1-x)\) on \([0,1]\) .

   Answer

   (a) The limit is \(0\) . On \([0,1]\) , \(\lVert f_n\rVert_\infty=1/n\to 0\) : uniform. On \(\mathbb R\) , \(\lVert f_n\rVert_\infty=\infty\) for every \(n\) : not uniform. (b) The limit is \(0\) at \(x=0\) and \(1\) for \(x>0\) . On \([0,1]\) the limit is discontinuous, so the convergence is not uniform, by Theorem 8.5. On \([1,2]\) , \(\lvert f_n(x)-1\rvert=1/(1+nx)\le 1/(1+n)\) : uniform. (c) The limit is \(0\) . The maximum of \(x^n(1-x)\) is at \(x=n/(n+1)\) and is less than \(1/(n+1)\) : uniform.

2. Prove directly from Definition 8.2 that \(f_n(x)=x+x^2/n\) converges uniformly to \(x\) on \([0,5]\) , and find the smallest \(N\) that works for \(\varepsilon=0.01\) . Show that the convergence is not uniform on \(\mathbb R\) .

   Solution

   Uniform convergence on \([0,5]\) . For \(x\in[0,5]\) ,

   \[ \lvert f_n(x)-x\rvert=\frac{x^2}{n}\le\frac{25}{n}. \]

   Let \(\varepsilon>0\) and choose \(N\) with \(N>25/\varepsilon\) . For \(n\ge N\) and every \(x\in[0,5]\) we get \(\lvert f_n(x)-x\rvert\le 25/n<\varepsilon\) . The index \(N\) was chosen without reference to \(x\) , so the convergence is uniform.

   The smallest \(N\) for \(\varepsilon=0.01\) . The bound \(25/n\) is attained, at \(x=5\) . So the condition “\(\lvert f_n(x)-x\rvert<0.01\) for all \(x\in[0,5]\) ” holds exactly when \(25/n<0.01\) , that is, \(n>2500\) . For \(n=2500\) the error at \(x=5\) equals \(0.01\) , which is not less than \(0.01\) . The smallest \(N\) is \(2501\) .

   Not uniform on \(\mathbb R\) . For each \(n\) , \(\sup_{x\in\mathbb R}x^2/n=\infty\) . Concretely, at the moving point \(x_n=n\) the error is \(n^2/n=n\) , which does not tend to \(0\) . By Proposition 8.3 the convergence is not uniform. The pointwise limit is the same on both sets; only the uniformity differs.

3. Let \(f_n(x)=nxe^{-nx}\) on \([0,\infty)\) . Find the pointwise limit, compute \(\lVert f_n\rVert_\infty\) , and show that the convergence is not uniform on \([0,\infty)\) but is uniform on \([a,\infty)\) for every \(a>0\) .

   Solution

   Pointwise limit. \(f_n(0)=0\) for all \(n\) . For \(x>0\) put \(t=nx\) ; then \(f_n(x)=te^{-t}\) , and since \(e^t\ge t^2/2\) for \(t>0\) we have \(te^{-t}\le 2/t=2/(nx)\to 0\) . So \(f_n\to 0\) pointwise on \([0,\infty)\) .

   The supremum. \(f_n'(x)=ne^{-nx}(1-nx)\) , which is positive for \(x<1/n\) and negative for \(x>1/n\) . So \(f_n\) increases on \([0,1/n]\) and decreases on \([1/n,\infty)\) , and

   \[ \lVert f_n\rVert_\infty=f_n(1/n)=e^{-1}\approx 0.3679\qquad\text{for every }n . \]

   This does not tend to \(0\) , so the convergence is not uniform on \([0,\infty)\) . Each \(f_n\) is the same hump \(te^{-t}\) , compressed horizontally by the factor \(n\) ; the hump moves towards \(0\) without getting lower.

   On \([a,\infty)\) . Let \(a>0\) . For \(n>1/a\) the point \(1/n\) lies to the left of \(a\) , so \(f_n\) is decreasing on \([a,\infty)\) and its supremum there is \(f_n(a)=nae^{-na}\) , which tends to \(0\) by the first part. Hence the convergence is uniform on \([a,\infty)\) . Removing any neighbourhood of \(0\) removes the place where the hump hides.

4. Prove that if \(f_n\to f\) and \(g_n\to g\) uniformly on \(S\) , then \(f_n+g_n\to f+g\) uniformly. Show that \(f_ng_n\) need not converge uniformly to \(fg\) by taking \(f_n(x)=g_n(x)=x+1/n\) on \(\mathbb R\) , and state a hypothesis under which it does.

   Hint

   For the sum use \(\lvert(f_n+g_n)-(f+g)\rvert\le\lVert f_n-f\rVert_\infty+\lVert g_n-g\rVert_\infty\) . For the product, \((x+1/n)^2-x^2=2x/n+1/n^2\) is unbounded in \(x\) for each \(n\) . The product converges uniformly if \(f\) and \(g\) are bounded: write \(f_ng_n-fg=f_n(g_n-g)+g(f_n-f)\) and note that \((f_n)\) is then uniformly bounded for large \(n\) .

5. Prove that a uniform limit of bounded functions is bounded. Give an example of a sequence of bounded functions on \((0,1]\) that converges pointwise to an unbounded function.

   Solution

   The claim. Let \(f_n\to f\) uniformly on \(S\) , each \(f_n\) bounded, say \(\lvert f_n\rvert\le B_n\) . By uniform convergence with \(\varepsilon=1\) there is \(N\) with \(\lvert f_N(x)-f(x)\rvert<1\) for all \(x\in S\) . Then for all \(x\) ,

   \[ \lvert f(x)\rvert\le\lvert f(x)-f_N(x)\rvert+\lvert f_N(x)\rvert<1+B_N . \]

   So \(f\) is bounded. One index \(N\) serves all points, which is where uniformity is used.

   The example. On \((0,1]\) let \(f_n(x)=\min\{n,\ 1/x\}\) . Each \(f_n\) is bounded by \(n\) . For fixed \(x\) , as soon as \(n\ge 1/x\) we have \(f_n(x)=1/x\) ; so \(f_n\to f\) pointwise with \(f(x)=1/x\) , which is unbounded on \((0,1]\) . By the first part the convergence cannot be uniform, and indeed \(\lVert f_n-f\rVert_\infty=\sup_{0<x<1/n}(1/x-n)=\infty\) .

   Pointwise limits can therefore destroy boundedness; this is one more property, besides continuity and integrals, that needs uniformity.

6. Let \(f_n:X\to\mathbb R\) be uniformly continuous functions on a metric space that converge uniformly to \(f\) . Prove that \(f\) is uniformly continuous.

   Solution

   Let \(\varepsilon>0\) . By uniform convergence choose \(N\) with \(\lvert f_N(x)-f(x)\rvert<\varepsilon/3\) for all \(x\in X\) . The single function \(f_N\) is uniformly continuous, so there is \(\delta>0\) with \(\lvert f_N(x)-f_N(y)\rvert<\varepsilon/3\) whenever \(d(x,y)<\delta\) . For such \(x,y\) ,

   \[ \lvert f(x)-f(y)\rvert\le\lvert f(x)-f_N(x)\rvert+\lvert f_N(x)-f_N(y)\rvert+\lvert f_N(y)-f(y)\rvert<\varepsilon . \]

   The number \(\delta\) depends only on \(\varepsilon\) (through \(N\) ), not on \(x\) or \(y\) . So \(f\) is uniformly continuous.

   This is the proof of Theorem 8.5 with “continuous at \(a\) ” replaced by “uniformly continuous”: the three-term splitting transfers to the limit whatever kind of control the approximant \(f_N\) has, provided the first and third terms are controlled uniformly.

7. Find the flaw: “Let \(0\le x<1\) and \(\varepsilon>0\) . Since \(x^n\to 0\) , choose \(N\) with \(x^N<\varepsilon\) . Then \(x^n<\varepsilon\) for all \(n\ge N\) . Hence \(x^n\to 0\) uniformly on \([0,1)\) .”

   Answer

   The index \(N\) was chosen after \(x\) was fixed, so it depends on \(x\) ; the argument proves pointwise convergence only. No single \(N\) works for all \(x\in[0,1)\) , because \(\sup_{0\le x<1}x^n=1\) for every \(n\) .

8. It is known that for each \(x\ge 0\) the sequence \((1+x/n)^n\) is increasing in \(n\) and converges to \(e^x\) . Use Dini’s theorem to conclude that the convergence is uniform on \([0,1]\) , checking each hypothesis.

   Hint

   Check: \([0,1]\) is compact; each \(f_n(x)=(1+x/n)^n\) is a polynomial, hence continuous; for each \(x\) the sequence is increasing in \(n\) (given); the pointwise limit \(e^x\) is continuous. Dini’s theorem in its increasing form (apply Theorem 8.6 to \(-f_n\) ) gives uniform convergence.

9. Let \(f_n(x)=nx(1-x^2)^n\) on \([0,1]\) . Find the pointwise limit, compute \(\int_0^1f_n\) , and deduce without computing a supremum that the convergence is not uniform. Then confirm this by finding \(\lVert f_n\rVert_\infty\) .

   Solution

   Pointwise limit. \(f_n(0)=0\) and \(f_n(1)=0\) . For \(0<x<1\) , \(f_n(x)=x\cdot nr^n\) with \(r=1-x^2\in(0,1)\) , and \(nr^n\to 0\) because \(\sum nr^n\) converges by the ratio test. So \(f_n\to 0\) pointwise on \([0,1]\) .

   Integrals. With the substitution \(u=1-x^2\) , \(du=-2x\,dx\) ,

   \[ \int_0^1nx(1-x^2)^n\,dx=\frac n2\int_0^1u^n\,du=\frac{n}{2(n+1)}\longrightarrow\frac12 . \]

   The integral of the limit function is \(0\) . If the convergence were uniform, Theorem 8.7 would force \(\int_0^1f_n\to 0\) . Since the integrals tend to \(\tfrac12\) , the convergence is not uniform.

   The supremum. \(f_n'(x)=n(1-x^2)^{n-1}\bigl(1-(2n+1)x^2\bigr)\) , which vanishes in \((0,1)\) only at \(x=1/\sqrt{2n+1}\) , a maximum. So

   \[ \lVert f_n\rVert_\infty=\frac{n}{\sqrt{2n+1}}\Bigl(\frac{2n}{2n+1}\Bigr)^n . \]

   By Bernoulli’s inequality \((1-\tfrac{1}{2n+1})^n\ge 1-\tfrac{n}{2n+1}\ge\tfrac12\) , so \(\lVert f_n\rVert_\infty\ge n/(2\sqrt{2n+1})\to\infty\) . For \(n=10\) the supremum is about \(1.340\) and for \(n=100\) about \(4.283\) .

10. Evaluate \(\lim_{n\to\infty}\int_0^1\dfrac{nx+1}{n+x}\,dx\) , justifying the interchange of limit and integral.

    Solution

    Let \(f_n(x)=(nx+1)/(n+x)\) on \([0,1]\) . For \(x\in[0,1]\) ,

    \[ f_n(x)-x=\frac{nx+1-nx-x^2}{n+x}=\frac{1-x^2}{n+x},\qquad\text{so}\qquad\lvert f_n(x)-x\rvert\le\frac1n . \]

    Hence \(\lVert f_n-x\rVert_\infty\le 1/n\to 0\) : the functions \(f_n\) , which are continuous and therefore integrable, converge uniformly on \([0,1]\) to \(f(x)=x\) . By Theorem 8.7,

    \[ \lim_{n\to\infty}\int_0^1\frac{nx+1}{n+x}\,dx=\int_0^1x\,dx=\frac12 . \]

    As a check, the integral can be computed exactly: \(\dfrac{nx+1}{n+x}=n-\dfrac{n^2-1}{n+x}\) , so it equals \(n-(n^2-1)\log\dfrac{n+1}{n}\) , which is about \(0.5643\) for \(n=10\) , \(0.5066\) for \(n=100\) , and \(0.5007\) for \(n=1000\) .

    The point of the exercise is the order of the steps. One does not need the exact integral, and for most integrands one could not find it. It is enough to identify the pointwise limit, bound the difference by a quantity free of \(x\) that tends to zero, and quote the theorem; the uniform bound \(1/n\) even gives the rate, since the integrals differ from \(\tfrac12\) by at most \(1/n\) .

11. Let \(f_n(x)=x^n/n\) on \([0,1]\) . Show that \(f_n\to 0\) uniformly, that \(f_n'\) converges pointwise, and that the limit of the derivatives is not the derivative of the limit at \(x=1\) . Which hypothesis of Theorem 8.8 fails?

    Answer

    \(\lVert f_n\rVert_\infty=1/n\to 0\) , so \(f_n\to 0\) uniformly and the limit has derivative \(0\) everywhere. The derivatives \(f_n'(x)=x^{n-1}\) converge pointwise to \(0\) for \(x<1\) and to \(1\) at \(x=1\) . At \(x=1\) the limit of the derivatives is \(1\) and the derivative of the limit is \(0\) . Theorem 8.8 requires \(f_n'\) to converge uniformly; here the limit of \(f_n'\) is discontinuous, so the convergence is not uniform on \([0,1]\) .

12. Use the M-test to prove that \(\sum_{n=1}^\infty 1/(n^2+x^2)\) converges uniformly on \(\mathbb R\) to a continuous function. Then prove that the geometric series \(\sum x^n\) does not converge uniformly on \((-1,1)\) , although it converges at every point.

    Solution

    The first series. For every real \(x\) and every \(n\ge 1\) , \(0<1/(n^2+x^2)\le 1/n^2\) , and \(\sum 1/n^2\) converges. By the Weierstrass M-test with \(M_n=1/n^2\) the series converges uniformly on \(\mathbb R\) . Each term is a continuous function of \(x\) , so each partial sum is continuous, and by Theorem 8.5 the sum is continuous on \(\mathbb R\) .

    The geometric series. For \(\lvert x\rvert<1\) the partial sums are \(s_N(x)=(1-x^{N+1})/(1-x)\) and the sum is \(s(x)=1/(1-x)\) , so

    \[ s(x)-s_N(x)=\frac{x^{N+1}}{1-x}. \]

    For fixed \(N\) this is unbounded on \((-1,1)\) : as \(x\to 1^-\) the numerator tends to \(1\) and the denominator to \(0\) . So \(\lVert s-s_N\rVert_\infty=\infty\) for every \(N\) , and the convergence is not uniform on \((-1,1)\) . A second argument: each \(s_N\) is bounded on \((-1,1)\) , and a uniform limit of bounded functions is bounded (Exercise 5), but \(s\) is not.

    On each smaller interval \([-r,r]\) with \(r<1\) the convergence is uniform, by the M-test with \(M_n=r^n\) ; this is the general pattern for power series.

13. Find the radius of convergence: (a) \(\sum n!\,x^n\) ; (b) \(\sum x^n/n^n\) ; (c) \(\sum 2^nx^n/n\) ; (d) \(\sum\bigl(2+(-1)^n\bigr)^nx^n\) .

    Answer

    (a) \(R=0\) : the ratio of consecutive terms is \((n+1)\lvert x\rvert\to\infty\) for \(x\ne 0\) . (b) \(R=\infty\) : the \(n\) th root of the \(n\) th term is \(\lvert x\rvert/n\to 0\) . (c) \(R=\tfrac12\) : the \(n\) th root of \(2^n/n\) tends to \(2\) . (d) \(R=\tfrac13\) : \(\lvert a_n\rvert^{1/n}\) is \(3\) for even \(n\) and \(1\) for odd \(n\) , so the limit superior is \(3\) .

14. Use the identities of Example 8.6 to find the sums \(\sum_{n=1}^\infty n/3^n\) and \(\sum_{n=1}^\infty n^2/3^n\) .

    Solution

    Both identities of Example 8.6 hold for \(\lvert x\rvert<1\) , and \(x=\tfrac13\) is in that range.

    First sum. \(\displaystyle\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\) gives

    \[ \sum_{n=1}^\infty\frac{n}{3^n}=\frac{1/3}{(2/3)^2}=\frac13\cdot\frac94=\frac34 . \]

    Second sum. \(\displaystyle\sum_{n=1}^\infty n^2x^n=\frac{x(1+x)}{(1-x)^3}\) gives

    \[ \sum_{n=1}^\infty\frac{n^2}{3^n}=\frac{(1/3)(4/3)}{(2/3)^3}=\frac49\cdot\frac{27}{8}=\frac32 . \]

    The justification is Theorem 8.11: inside the interval of convergence a power series may be differentiated term by term, so the identities obtained from the geometric series are valid for every \(x\) with \(\lvert x\rvert<1\) , not merely formal. Partial sums confirm the values: through \(n=30\) they are \(0.749999\ldots\) and \(1.499999\ldots\) .

15. Use the series for \(\log(1+x)\) to approximate \(\log(1.1)\) by its first three terms, and bound the error using the alternating series estimate of Theorem 3.15.

    Solution

    For \(\lvert x\rvert<1\) , \(\log(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots\) . At \(x=0.1\) the first three terms give

    \[ 0.1-0.005+0.000333\ldots=0.095333\ldots \]

    The series at \(x=0.1\) is alternating with terms \(b_n=0.1^n/n\) , which decrease to \(0\) . By Theorem 3.15 the error after three terms is at most the fourth term:

    \[ \Bigl\lvert\log(1.1)-0.095333\ldots\Bigr\rvert\le\frac{0.1^4}{4}=0.000025 , \]

    and the sum lies between the third and fourth partial sums: \(0.095308<\log(1.1)<0.095334\) . The true value is \(0.095310\) to six places.

    Compare Example 3.6, where a hundred terms of the same series at \(x=1\) gave two decimal places. At \(x=0.1\) three terms give four. The rate of convergence of a power series depends on how far the point is from the edge of the interval of convergence.

16. Define \(C(x)=\sum_{n=0}^\infty(-1)^nx^{2n}/(2n)!\) and \(S(x)=\sum_{n=0}^\infty(-1)^nx^{2n+1}/(2n+1)!\) . Show that both series have infinite radius of convergence, that \(S'=C\) and \(C'=-S\) , and that \(S(x)^2+C(x)^2=1\) for all \(x\) .

    Hint

    For fixed \(x\) the ratio of consecutive terms of either series is \(x^2\) divided by a product of two consecutive integers, which tends to \(0\) ; so both converge for all \(x\) . Differentiate term by term using Theorem 8.11. For the last part, \((S^2+C^2)'=2SC-2CS=0\) , so \(S^2+C^2\) is constant, equal to its value \(0+1\) at \(x=0\) .

17. Suppose \(\sum a_nx^n\) and \(\sum b_nx^n\) both converge on \((-r,r)\) , \(r>0\) , and have the same sum there. Prove that \(a_n=b_n\) for all \(n\) .

    Hint

    By Theorem 8.11 the coefficients are determined by the sum: \(a_n=f^{(n)}(0)/n!=b_n\) , where \(f\) is the common sum on \((-r,r)\) .

18. In the metric space \(C[0,1]\) with the supremum metric, prove that \(\{f:f(0)=0\}\) is a closed set, and that the sequence \(f_n(x)=x^n\) lies in the closed unit ball and has no convergent subsequence. What does this say about that ball?

    Solution

    The set is closed. Let \(A=\{f\in C[0,1]:f(0)=0\}\) , and let \(f_k\in A\) converge to \(f\) in \(C[0,1]\) , that is, \(\lVert f_k-f\rVert_\infty\to 0\) . Then

    \[ \lvert f(0)\rvert=\lvert f(0)-f_k(0)\rvert\le\lVert f-f_k\rVert_\infty\to 0 , \]

    so \(f(0)=0\) and \(f\in A\) . By Theorem 4.12, \(A\) is closed.

    The sequence of powers. \(f_n(x)=x^n\) is continuous with \(\lVert f_n\rVert_\infty=1\) , so it lies in the closed unit ball of \(C[0,1]\) . Suppose a subsequence \(f_{n_k}\) converged in \(C[0,1]\) to some \(g\) . Convergence in the supremum metric is uniform convergence, which implies pointwise convergence; so \(g(x)=\lim x^{n_k}\) , which is \(0\) for \(x<1\) and \(1\) for \(x=1\) . That function is not continuous, contradicting \(g\in C[0,1]\) . Hence no subsequence converges.

    Conclusion. The closed unit ball of \(C[0,1]\) is closed and bounded, and by Theorem 4.17 it is not compact. The Heine–Borel theorem is a theorem about \(\mathbb R^n\) and fails in this space of functions, as it did for the space of bounded sequences in Example 4.5.

19. Compute the Bernstein polynomial \(B_nf\) for \(f(x)=x^2\) in closed form, using Lemma 8.13, and find \(\lVert B_nf-f\rVert_\infty\) exactly.

    Solution

    For \(f(x)=x^2\) , \(f(k/n)=k^2/n^2\) , and \(k^2=k(k-1)+k\) . By the identities in the proof of Lemma 8.13, \(\sum_kk(k-1)p_{n,k}(x)=n(n-1)x^2\) and \(\sum_kkp_{n,k}(x)=nx\) . Therefore

    \[ B_nf(x)=\frac{1}{n^2}\bigl(n(n-1)x^2+nx\bigr)=x^2+\frac{x(1-x)}{n}. \]

    So \(B_nf(x)-f(x)=x(1-x)/n\ge 0\) on \([0,1]\) , largest at \(x=\tfrac12\) , and

    \[ \lVert B_nf-f\rVert_\infty=\frac{1}{4n}. \]

    For example \(\lVert B_{10}f-f\rVert_\infty=0.025\) . Note that \(B_nf\) is a polynomial of degree \(2\) that is not equal to \(x^2\) : the Bernstein polynomial of a polynomial is not the polynomial itself. The convergence is uniform, at the slow rate \(1/n\) , in agreement with Theorem 8.14 and with the remark in Example 8.8 about efficiency.

20. Let \(f\) be continuous on \([0,1]\) with \(\int_0^1f(x)\,x^n\,dx=0\) for every integer \(n\ge 0\) . Prove that \(f\) is identically zero.

    Solution

    By linearity of the integral, the hypothesis gives \(\int_0^1f(x)p(x)\,dx=0\) for every polynomial \(p\) .

    By the Weierstrass approximation theorem there are polynomials \(p_k\) with \(\lVert p_k-f\rVert_\infty\to 0\) . Let \(M=\lVert f\rVert_\infty\) , finite because \(f\) is continuous on a compact interval. Then

    \[ \lVert fp_k-f^2\rVert_\infty\le M\,\lVert p_k-f\rVert_\infty\to 0 , \]

    so \(fp_k\to f^2\) uniformly on \([0,1]\) . By Theorem 8.7,

    \[ \int_0^1f^2=\lim_{k\to\infty}\int_0^1fp_k=0 . \]

    The function \(f^2\) is continuous and nonnegative with integral \(0\) , so \(f^2\) is identically zero by Exercise 7.7. Hence \(f=0\) .

    The numbers \(\int_0^1f(x)x^n\,dx\) are called the moments of \(f\) ; the exercise says that a continuous function on \([0,1]\) is determined by its moments.

21. Suppose a sequence of polynomials converges uniformly on all of \(\mathbb R\) to a function \(f\) . Prove that \(f\) is a polynomial. Why does this not contradict Theorem 8.14?

    Hint

    By the uniform Cauchy criterion there is \(N\) with \(\lVert p_n-p_N\rVert_\infty<1\) on \(\mathbb R\) for \(n\ge N\) . A polynomial that is bounded on \(\mathbb R\) is constant, so \(p_n=p_N+c_n\) with constants \(c_n\) , which converge. Hence \(f=p_N+\lim c_n\) . Theorem 8.14 concerns compact intervals, where polynomials of growing degree stay bounded.

22. Prove that \(f_n\to f\) uniformly on \(S\) if and only if \(f_n(x_n)-f(x_n)\to 0\) for every sequence \((x_n)\) of points of \(S\) .

    Hint

    If \(f_n\to f\) uniformly, then \(\lvert f_n(x_n)-f(x_n)\rvert\le\lVert f_n-f\rVert_\infty\to 0\) . If the convergence is not uniform, there are \(\varepsilon_0>0\) and infinitely many indices \(n\) with a point \(x_n\) such that \(\lvert f_n(x_n)-f(x_n)\rvert\ge\varepsilon_0\) ; choose \(x_n\) arbitrarily for the other indices.