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陶哲轩实分析11.10及习题-Analysis I 11.10

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Exercise 11.10.1. Prove Proposition 11.10.1.
Proposition 11.10.1 (Integration by parts formula). Let I=[a,b], and let F:[a,b]\to\mathbf R and G:[a,b]\to\mathbf R be differentiable functions on [a,b] such that F' and G' are Riemann integrable on I. Then we have

\displaystyle{\int_{[a,b]}FG'=F(b)G(b)-F(a)G(a)-\int_{[a,b]}F'G}.

Solution: As F and G are differentiable, they are continuous on [a,b] and thus Riemann integrable by Corollary 11.5.2, then FG' and F'G are Riemann integrable by Theorem 11.4.5. Notice that (FG)'=F' G+FG' and F'G+FG' is Riemann integrable on [a,b], we can use the second fundamental theorem of calculus to have

\displaystyle{\int_{[a,b]}(F'G+FG')=(FG)(b)-(FG)(a)}

By some simple calculation we can have the final result.

\blacksquare

Exercise 11.10.2. fill in the gaps marked (why?) in the proof of Lemma 11.10.5.
Lemma 11.10.5 (Change of variables formula I). Let [a,b] be a closed interval, and let \phi:[a,b]\to [\phi (a),\phi (b)] be a continuous monotone increasing function. Let f:[\phi (a),\phi (b)]\to\mathbf R be a piecewise constant function on [\phi (a),\phi (b)]. Then f\circ \phi:[a,b]\to\mathbf R is also piecewise constant on [a,b], and

\displaystyle{\int_{[a,b]}f\circ \phi d\phi =\int_{[\phi (a),\phi (b)]}f}.

Solution: I will prove Lemma 11.10.5 completely.
Let \mathbf P be a partition of [{\phi}(a),{\phi}(b)] such that f is piecewise constant with respect to \mathbf P. we may assume that \mathbf P does not contain the empty set. For each J\in \mathbf P, let c_J be the constant value of f on J, thus

\displaystyle{\int_{[{\phi}(a),{\phi}(b)]} f=\sum_{J\in \mathbf P}c_J |J|}

For each interval J, let {\phi}^{-1} (J):=\{x\in [a,b]:{\phi}(x)\in J\}. Then given any x,y\in {\phi}^{-1} (J),x<y, we have {\phi}(x),{\phi}(y)\in J by the definition of {\phi}^{-1} (J), and {\phi}(x)\leq {\phi}(y) since {\phi} is monotone increasing, and \forall z\in [x,y], we have {\phi}(x)\leq {\phi}(z)\leq {\phi}(y) again by {\phi} is monotone increasing, this means {\phi}(z)\in J since J is an interval, so {\phi}^{-1} (J) is connected, and thus is an interval. Furthermore, if x\in {\phi}^{-1} (J), then {\phi}(x)\in J and f({\phi}(x))=c_J, thus c_J is the constant value of f\circ {\phi} on {\phi}^{-1}(J).
Thus if we define \mathbf Q:=\{{\phi}^{-1}(J):J\in \mathbf P\}, then for any c\in [a,b], {\phi}(c)\in [{\phi}(a),{\phi}(b)], so there is a J\in \mathbf P such that {\phi}(c)\in J, which means c\in {\phi}^{-1}(J). Assume we can find two sets Q_1,Q_2\in \mathbf Q s.t. c\in Q_1\cap Q_2, then it means we can find J_1,J_2\in \mathbf P s.t. {\phi}(c)\in J_1\cap J_2, contradict the fact that \mathbf P is a partition of [{\phi}(a),{\phi}(b)]. We can conclude Q is a partition of [a,b].
Given any Q\in \mathbf Q, we have Q={\phi}^{-1}(J) for some J\in \mathbf P, so c_J is the constant value of f\circ {\phi} on this Q, which means f\circ {\phi} piecewise constant with respect to \mathbf Q. Thus

\displaystyle{\int_{[a,b]}f\circ {\phi} d{\phi}=\int_{[\mathbf Q]}f\circ {\phi} d{\phi}=\sum_{J\in \mathbf P}c_J {\phi}[{\phi}^{-1}(J)] }

To calculate {\phi}[{\phi}^{-1}(J)], we let J\in \mathbf P, then J is an interval with endpoints a<b, and |J|=b-a, as {\phi} is monotone increasing, we can find a',b'\in {\phi}^{-1} (J), s.t. {\phi}(a' )=a,{\phi}(b')=b, and {\phi}^{-1}(J) is an interval which has endpoints a',b', thus {\phi}[{\phi}^{-1}(J)]={\phi}(b' )-{\phi}(a')=b-a=|J|, and the claim follows.

\blacksquare

Exercise 11.10.3. Let a<b be real numbers, and let f:[a,b]\to\mathbf R be Riemann integrable function. Let g:[-b,-a]\to\mathbf R be defined by g(x):=f(-x). Show that g is also Riemann integrable, and \int_{[-b,-a]}g=\int_{[a,b]}f.

Solution: For \forall {\varepsilon}>0, we can find a partition of [a,b], namely \mathbf P, and piecewise constant function \overline{f} which majorizes f and \underline{f} which minorizes f, both with respect to \mathbf P, such that

\displaystyle{\int_{[a,b]}f-{\varepsilon}<\int_{[a,b]}\underline{f}\leq \int_{[a,b]}\overline{f}<\int_{[a,b]}f+{\varepsilon}}

Now we define \overline{g}(x)=\overline{f}(-x) and \underline{g}(x)=\underline{f}(-x) on [-b,-a], then it is easy to see that \overline{g} majorizes g and \underline{g} minorizes g. For any J\in \mathbf P, we define K_J=\{-x:x\in J\}, then |K_J |=|J|, and \mathbf P'=\{K_J:J\in \mathbf P\} is a partition of [-b,-a], and the constant value of \overline{g} on K_J is the same as the constant value of \overline{f} on J, the constant value of \underline{g} on K_J is the same as the constant value of \underline{f} on J, so we have

\displaystyle{\int_{[-b,-a]}\overline{g}=p.c.\int_{[-b,-a]}\overline{g}=\sum_{K_J\in \mathbf P'}c_J |K_J|=\sum_{J\in \mathbf P}c_J |J| =\int_{[a,b]}\overline{f}}

Similarly we have

\displaystyle{\int_{[-b,-a]}\underline{g}=\int_{[a,b]}\underline{f}}

So we have

\displaystyle{\int_{[a,b]}f-{\varepsilon}<\int_{[-b,-a]}\underline{g}\leq \underline{\int}_{[-b,-a]}g\leq \overline{\int}_{[-b,-a]}g\leq \int_{[-b,-a]}\overline{g}<\int_{[a,b]}f+{\varepsilon}}

and the conclusion follows.

\blacksquare

Exercise 11.10.4. What is the analogue of Proposition 11.10.7 when \phi is monotone decreasing instead of monotone increasing?

Solution: Let [a,b] be a closed interval, and let {\phi}:[a,b]\to [{\phi}(b),{\phi}(a)] be a differentiable monotone decreasing function such that {\phi}' is Riemann integrable. Let f: [{\phi}(b),{\phi}(a)]\to \mathbf R be a Riemann integrable function on [{\phi}(b),{\phi}(a)]. Then (f\circ {\phi}) {\phi}':[a,b]\to \mathbf R is Riemann integrable on [a,b] and

\displaystyle{\int_{[a,b]}(f\circ {\phi}) {\phi}'=-\int_{[{\phi}(b),{\phi}(a)]}f}

\blacksquare

陶哲轩实分析11.9及习题-Analysis I 11.9

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Exercise 11.9.1. Let f:[0,1]\to\mathbf R be the function in Exercise 9.8.5. Show that for every rational number q\in \mathbf Q\cap [0,1], the function F:[0,1]\to\mathbf R defined by the formula F(x):=\int_0^xf(y)dy is not differentiable at q.

Solution: Since f is monotonic increasing, f is Riemann ingegrable and F is well-defined.
We assume F(x) is differentiable at some q\in Q\cap [0,1], then if q\neq 0,q\neq 1, F is differentiable at q iff we have

\displaystyle{\lim_{x\to q;x\in [0,q)}\frac{F(x)-F(q)}{x-q}=\lim_{x\to q;x\in (q,1]}\frac{F(x)-F(q)}{x-q}}

From Exercise 9.8.5 we know that f is not continuous at q and if we locate q=q(n) for some n\in \mathbf N as in Exercise 9.8.5, we shall have f(x)\geq f(q)+2^{-n},x>q, so

\lim\limits_{x\to q;x\in [0,q)}\dfrac{F(x)-F(q)}{x-q}=\lim\limits_{x\to q;x\in [0,q)}\dfrac{\int_x^qf(y)dy}{x-q}\leq \lim\limits_{x\to q;x\in [0,q)}\dfrac{\int_x^qf(q)dy}{x-q}=f(q) \\ \lim\limits_{x\to q;x\in (q,1]}\dfrac{F(x)-F(q)}{x-q}=\lim\limits_{x\to q;x\in (q,1]}\dfrac{\int_q^xf(y)dy}{x-q}\geq \lim\limits_{x\to q;x\in (q,1]}\dfrac{\int_q^x(f(q)+2^{-n})dy}{x-q}=f(q)+2^{-n}

thus

\displaystyle{\lim_{x\to q;x\in (q,1]}\frac{F(x)-F(q)}{x-q}-\lim_{x\to q;x\in [0,q)}\frac{F(x)-F(q)}{x-q}\geq 2^{-n}}

contradicts the result that the two limits are equal.
If q=0 or q=1, then f is discontinuous at 0 or 1 when restricted on [0,1], due to Exercise 9.8.5, thus F is not differentiable, as the limit \lim_{x\to 0;x\in (0,1]}\frac{F(x)}{x} and \lim_{x\to 1;x\in [0,1)}\frac{F(x)-F(1)}{x-1} do not exist.

\blacksquare

Exercise 11.9.2. Prove Lemma 11.9.5.
Lemma 11.9.5 Let I be a bounded interval, and let f:I\to\mathbf R be a function. Let F:I\to\mathbf R and G:I\to\mathbf R be two antiderivatives of f. Then there exists a real number C such that F(x)=G(x)+C for all x\in I.

Solution: When I is empty or a single point then the statement is trivial. Now suppose I is a nonempty interval, choose some x_0\in I, then for any x\in I,x\neq x_0, the function F-G is differentiable on [x_0,x] or [x,x_0], thus we have some {\xi} between x and x_0, s.t.

\displaystyle{\frac{(F-G)(x)-(F-G)(x_0)}{x-x_0}=(F-G)'({\xi})=f({\xi})-f({\xi})=0}

So we have

\displaystyle{(F-G)(x)=F(x)-G(x)=F(x_0 )-G(x_0 )}

This means if we let C=F(x_0)-G(x_0), we will have F(x)=G(x)+C,\forall x\in I,x\neq x_0, since we also have F(x_0 )=G(x_0 )+C valid, the conclusion is true.
Alternatively, we can use the second Fundamental theorem of calculus, since

\displaystyle{(F-G)'(x)=0,\quad \forall x\in I}

the function F-G:I\to \mathbf R is an antiderivative of the function g(x)=0,x\in I. Since g is constant, it is Riemann integrable, so choose some x_0\in I, then for any x\in I,x\neq x_0, we can use the second Fundamental theorem of calculus on [x_0,x] or [x,x_0] and have

\displaystyle{\int_{[x_0,x]}g=0=(F-G)(x)-(F-G)(x_0)}

or

\displaystyle{\int_{[x,x_0]}g=0=(F-G)(x_0 )-(F-G)(x)}

In either case we have (F-G)(x_0)=(F-G)(x), This means if we let C=F(x_0)-G(x_0), we will have F(x)=G(x)+C,\forall x\in I,x\neq x_0. Since we also have F(x_0)=G(x_0)+C valid, the conclusion is true.

\blacksquare

Exercise 11.9.3. Let a<b be real numbers, and let f:[a,b]\to\mathbf R be a monotone increasing function. Let F:[a,b]\to\mathbf R be the fucntion F(x):=\int_{[a,x]}f. Let x_0 be an element of [a,b]. Show that F is differentiable at x_0 if and only if f is continuous at x_0.

Solution: Since f is monotone increasing, f is Riemann integrable, thus if f is continuous at x_0, then F is differentiable at x_0 by the first Fundamental theorem of calculus.
Conversely we let F is differentiable at x_0 and assume f is not continuous at x_0, then we shall have f(x_0-)0, s.t.

\displaystyle{f(x_0-)<{\eta}-{\varepsilon}<{\eta}<{\eta}+{\varepsilon}<f(x_0+)}

As f is monotone increasing, we know f(x)<{\eta}-{\varepsilon} if x<x_0 and f(x)>{\eta}+{\varepsilon} if x>x_0. Then for y>x_0, we have

\displaystyle{F(y)-F(x_0 )=\int_{[x_0,y]}f>\int_{[x_0,y]}({\eta}+{\varepsilon}) =({\eta}+{\varepsilon})(y-x_0 )}

And for y<x_0, we have

\displaystyle{F(x_0 )-F(y)=\int_{[y,x_0]}f<\int_{[y,x_0]}({\eta}-{\varepsilon}) =({\eta}-{\varepsilon})(x_0-y)}

So we have

\lim\limits_{y\to x_0+}\dfrac{F(y)-F(x_0)}{y-x_0}\geq {\eta}+{\varepsilon}>{\eta} \\ \lim\limits_{y\to x_0-}\dfrac{F(y)-F(x_0)}{y-x_0}=\lim\limits_{y\to x_0-}\dfrac{F(x_0)-F(y)}{x_0-y}\leq {\eta}-{\varepsilon}<{\eta}

thus F'(x_0+)>F'(x_0-), a contradiction to F being differentiable at x_0.

\blacksquare

陶哲轩实分析10.5及习题-Analysis I 10.5

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洛必达法则，计算极为好使，但使用需小心。

Exercise 10.5.1. Prove Proposition 10.5.1.
Proposition 10.5.1 (L’Hôpital’s rule I). Let X be a subset of \mathbf R, let f:X\to\mathbf R and g:X\to\mathbf R be functions, and let x_0\in X be a limit point of X. Suppose that f(x_0)=g(x_0)=0, that f and g are both differentiable at x_0, but g'(x_0)\neq 0. Then there exists a \delta >0 such that g(x)\neq 0 for all x\in (X\cap (x_0-\delta,x_0+\delta))-\{x_0\}, and

\lim\limits_{x\to x_0;x\in (X\cap (x_0-\delta,x_0+\delta))-\{x_0\}}\dfrac{f(x)}{g(x)}=\dfrac{f'(x_0)}{g'(x_0)}

Solution: By Proposition 10.1.7, since g'(x_0 )\neq 0, we can find a {\delta}>0 such that for all x\in X and |x-x_0 |\leq {\delta}, we have

|g(x)-g(x_0 )-g'(x_0 )(x-x_0 )|\leq \dfrac{|g'(x_0 )|}{2} |x-x_0 |

Using triangle inequality and g(x_0 )=0, we further get

\dfrac{|g'(x_0 )|}{2} |x-x_0 |\leq |g(x)|

Thus if x\in \left(X\cap (x_0-{\delta},x_0+{\delta})\right)-\{x_0 \}, we shall have |g(x)|>0, or g(x)\neq 0. In this case we can safely use Proposition 9.3.14 to have

\begin{aligned}\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} } \dfrac{f(x)}{g(x)}&=\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} } \frac{f(x)-f(x_0)}{g(x)-g(x_0 )}\\&=\dfrac{\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} }\dfrac{f(x)-f(x_0 )}{x-x_0 }}{\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} }\dfrac{g(x)-g(x_0)}{x-x_0 }} \\&=\dfrac{f'(x_0)}{g'(x_0 ) }\end{aligned}

\blacksquare

Exercise 10.5.2. Explain why Exercise 1.2.12 does not contradict either of the propositions in this section.

Solution: In the first example, we have g(x)=1+x\nrightarrow 0 if x\to 0.
In the second example, notice that to use the L’Hôpital’s rule, we need first the limit

\lim\limits_{x\to a;x\in (a,b]}\dfrac{f'(x)}{g'(x)}

exist, then we can safely use the rule, but in this example, the limit

\lim\limits_{x\to 0+;x\in (0,{\infty}) }\dfrac{(x^2 \sin (x^{-4}))'}{x'}

does not exist, due to the calculations in the textbook. So it’s not safe to use the L’Hôpital’s rule.

\blacksquare