Tag Archives: real-analysis

A differentiable real function with unbounded derivative around zero

Consider the real function defined on Rf(x)={0for x=0x2sin1x2for x0

f is continuous and differentiable on R{0}. For xR we have |f(x)|x2, which implies that f is continuous at 0. Also |f(x)f(0)x|=|xsin1x2||x| proving that f is differentiable at zero with f(0)=0. The derivative of f for x0 is f(x)=2xsin1x2=g(x)2xcos1x2=h(x) On the interval (1,1), g(x) is bounded by 2. However, for ak=1kπ with kN we have h(ak)=2kπ(1)k which is unbounded while limkak=0. Therefore f is unbounded in all neighborhood of the origin.

A Riemann-integrable map that is not regulated

For a Banach space X, a function f:[a,b]X is said to be regulated if there exists a sequence of step functions φn:[a,b]X converging uniformly to f.

One can prove that a regulated function f:[a,b]X is Riemann-integrable. Is the converse true? The answer is negative and we provide below an example of a Riemann-integrable real function that is not regulated. Let’s first prove following theorem.

THEOREM A bounded function f:[a,b]R that is (Riemann) integrable on all intervals [c,b] with a<c<b is integrable on [a,b].

PROOF Take M>0 such that for all x[a,b] we have |f(x)|<M. For ϵ>0, denote c=inf(a+ϵ4M,b+ba2). As f is supposed to be integrable on [c,b], one can find a partition P: c=x1<x2<<xn=b such that 0U(f,P)L(f,P)<ϵ2 where L(f,P),U(f,P) are the lower and upper Darboux sums. For the partition P: a=x0<c=x1<x2<<xn=b, we have 0U(f,P)L(f,P)2M(ca)+(U(f,P)L(f,P))<2Mϵ4M+ϵ2=ϵ We now prove that the function f:[0,1][0,1] defined by f(x)={1 if x{2k ; kN}0otherwise is Riemann-integrable (that follows from above theorem) and not regulated. Let's prove it. If f was regulated, there would exist a step function g such that |f(x)g(x)|<13 for all x[0,1]. If 0=x0<x1<<xn=1 is a partition associated to g and c1 the value of g on the interval (0,x1), we must have |1c1|<13 as f takes (an infinite number of times) the value 1 on (0,x1). But f also takes (an infinite number of times) the value 0 on (0,x1). Hence we must have |c1|<13. We get a contradiction as those two inequalities are not compatible.

A discontinuous midpoint convex function

Let’s recall that a real function f:RR is called convex if for all x,yR and λ[0,1] we have f((1λ)x+λy)(1λ)f(x)+λf(y) f is called midpoint convex if for all x,yR f(x+y2)f(x)+f(y)2 One can prove that a continuous midpoint convex function is convex. Sierpinski proved the stronger theorem, that a real-valued Lebesgue measurable function that is midpoint convex will be convex.

Can one find a discontinuous midpoint convex function? The answer is positive but requires the Axiom of Choice. Why? Because Robert M. Solovay constructed a model of Zermelo-Fraenkel set theory (ZF), exclusive of the axiom of choice where all functions are Lebesgue measurable. Hence convex according to Sierpinski theorem. And one knows that convex functions defined on R are continuous.

Referring to my previous article on the existence of discontinuous additive map, let’s use a Hamel basis B=(bi)iI of R considered as a vector space on Q. Take i1I, define f(i1)=1 and f(i)=0 for iI{i1} and extend f linearly on R. f is midpoint convex as it is linear. As the image of R under f is Q, f is discontinuous as explained in the discontinuous additive map counterexample.

Moreover, f is unbounded on all open real subsets. By linearity, it is sufficient to prove that f is unbounded around 0. Let’s consider i1i2I. G=bi1Z+bi2Z is a proper subgroup of the additive R group. Hence G is either dense of discrete. It cannot be discrete as the set of vectors {b1,b2} is linearly independent. Hence G is dense in R. Therefore, one can find a non vanishing sequence (xn)nN=(q1nbi1+q2nbi2)nN (with (q1n,q2n)Q2 for all nN) converging to 0. As {b1,b2} is linearly independent, this implies |q1n|,|q2n|n+ and therefore limn|f(xn)|=limn|f(q1nbi1+q2nbi2)|=limn|q1n|=.

A positive smooth function with all derivatives vanishing at zero

Let’s consider the set C(R) of real smooth functions, i.e. functions that have derivatives of all orders on R.

Does a positive function fC(R) with all derivatives vanishing at zero exists?

Such a map f cannot be expandable in power series around zero, as it would vanish in a neighborhood of zero. However, the answer to our question is positive and we’ll prove that f(x)={e1x2ifx00ifx=0 provides an example.

f is well defined and positive for x0. As limx01x2=, we get limx0f(x)=0 proving that f is continuous on R. Let’s prove by induction that for x0 and nN, f(n)(x) can be written as f(n)(x)=Pn(x)x3ne1x2 where Pn is a polynomial function. The statement is satisfied for n=1 as f(x)=2x3e1x2. Suppose that the statement is true for n then f(n+1)(x)=[Pn(x)x3n3nPn(x)x3n+1+2Pn(x)x3n+3]e1x2 hence the statement is also true for n+1 by taking Pn+1(x)=x3Pn(x)3nx2Pn(x)+2Pn(x). Which concludes our induction proof.

Finally, we have to prove that for all nN, limx0f(n)(x)=0. For that, we use the power expansion of the exponential map ex=n=0xnn!. For x0, we have |x|3ne1x2|x|3n(2n)!|x|4n=1(2n)!|x|n Therefore limx0|x|3ne1x2= and limx0f(n)(x)=0 as f(n)(x)=Pn(x)x3ne1x2 with Pn a polynomial function.

Counterexample around L’Hôpital’s rule

Let us consider two differentiable functions f and g defined in an open interval (a,b), where b might be . If
limxbf(x)=limxbg(x)= and if g(x)0 in some interval (c,b), then a version of l’Hôpital’s rule states that limxbf(x)g(x)=L implies limxbf(x)g(x)=L.

We provide a counterexample when g vanishes in all neighborhood of b. The counterexample is due to the Austrian mathematician Otto Stolz.

We take (0,) for the interval (a,b) and {f(x)=x+cosxsinxg(x)=esinx(x+cosxsinx) which derivatives are {f(x)=2cos2xg(x)=esinxcosx(x+cosxsinx+2cosx) We have limxf(x)g(x)=limx2cosxesinx(x+cosxsinx+2cosx)=0, however f(x)g(x)=1esinx doesn’t have any limit at as it oscillates between 1e and e.

On limit at infinity of functions and their derivatives

We consider continuously differentiable real functions defined on (0,) and the limits limxf(x) and limxf(x).

A map f such that limxf(x)= and limxf(x)=0

Consider the map f:xx. It is clear that limxf(x)=. As f(x)=12x, we have as announced limxf(x)=0

A bounded map g having no limit at infinity such that limxg(x)=0

One idea is to take an oscillating map whose wavelength is increasing to . Let’s take the map g:xcosx. g doesn’t have a limit at as for nN, we have g(n2π2)=cosnπ=(1)n. However, the derivative of g is g(x)=sinx2x, and as |g(x)|12x for all x(0,), we have limxg(x)=0.

Limit points of real sequences

Let’s start by recalling an important theorem of real analysis:

THEOREM. A necessary and sufficient condition for the convergence of a real sequence is that it is bounded and has a unique limit point.

As a consequence of the theorem, a sequence having a unique limit point is divergent if it is unbounded. An example of such a sequence is the sequence un=n2(1+(1)n), whose initial values are 0,1,0,2,0,3,0,4,0,5,6, (un) is an unbounded sequence whose unique limit point is 0.

Let’s now look at sequences having more complicated limit points sets.

A sequence whose set of limit points is the set of natural numbers

Consider the sequence (vn) whose initial terms are 1,1,2,1,2,3,1,2,3,4,1,2,3,4,5, (vn) is defined as follows vn={1 for n=1nk(k+1)2 for k(k+1)2<n(k+1)(k+2)2 (vn) is well defined as the sequence (k(k+1)2)kN is strictly increasing with first term equal to 1. (vn) is a sequence of natural numbers. As N is a set of isolated points of R, we have VN, where V is the set of limit points of (vn). Conversely, let’s take mN. For k+1m, we have k(k+1)2+m(k+1)(k+2)2, hence uk(k+1)2+m=m which proves that m is a limit point of (vn). Finally the set of limit points of (vn) is the set of natural numbers.

Continue reading Limit points of real sequences

Continuity versus uniform continuity

We consider real-valued functions.

A real-valued function f:IR (where I is an interval) is continuous at x0I when: (ϵ>0)(δ>0)(xI)(|xx0|δ|f(x)f(x0)|ϵ). When f is continuous at all xI, we say that f is continuous on I.

f:IR is said to be uniform continuity on I if (ϵ>0)(δ>0)(x,yI)(|xy|δ|f(x)f(y)|ϵ).

Obviously, a function which is uniform continuous on I is continuous on I. Is the converse true? The answer is negative.

An (unbounded) continuous function which is not uniform continuous

The map f:RRxx2 is continuous. Let’s prove that it is not uniform continuous. For 0<x<y we have |f(x)f(y)|=y2x2=(yx)(y+x)2x(yx) Hence for yx=δ>0 and x=1δ we get
|f(x)f(y)|2x(yx)=2>1 which means that the definition of uniform continuity is not fulfilled for ϵ=1.

For this example, the function is unbounded as limxx2=. Continue reading Continuity versus uniform continuity

No minimum at the origin but a minimum along all lines

We look here at an example, from the Italian mathematician Giuseppe Peano of a real function f defined on R2. f is having a local minimum at the origin along all lines passing through the origin, however f does not have a local minimum at the origin as a function of two variables.

The function f is defined as follows
f:R2R(x,y)f(x,y)=3x44x2y+y2 One can notice that f(x,y)=(y3x2)(yx2). In particular, f is strictly negative on the open set U={(x,y)R2 : x2<y<3x2}, vanishes on the parabolas y=x2 and y=3x2 and is strictly positive elsewhere. Consider a line D passing through the origin. If D is different from the coordinate axes, the equation of D is y=λx with λ>0. We have f(x,λx)=x2(λ3x)(λx). For x(,λ3){0}, f(x,λx)>0 while f(0,0)=0 which proves that f has a local minimum at the origin along the line Dyλx=0. Along the x-axis, we have f(x,0)=3x4 which has a minimum at the origin. And finally, f also has a minimum at the origin along the y-axis as f(0,y)=y2.

However, along the parabola Py=2x2 we have f(x,2x2)=x4 which is strictly negative for x0. As P is passing through the origin, f assumes both positive and negative values in all neighborhood of the origin.

This proves that f does not have a minimum at (0,0).

Counterexamples around Fubini’s theorem

We present here some counterexamples around the Fubini theorem.

We recall Fubini’s theorem for integrable functions:
let X and Y be σ-finite measure spaces and suppose that X×Y is given the product measure. Let f be a measurable function for the product measure. Then if f is X×Y integrable, which means that X×Y|f(x,y)|d(x,y)<, we have X(Yf(x,y)dy)dx=Y(Xf(x,y)dx)dy=X×Yf(x,y)d(x,y) Let's see what happens when some hypothesis of Fubini's theorem are not fulfilled. Continue reading Counterexamples around Fubini’s theorem