Existence of the limit at infinity for a function that is integrable on the half line

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1 Rose-Hulman Undergraduate Mathematics Journal Volume 14 Issue 1 Article 1 Existence of the limit at infinity for a function that is integrable on the half line James Patrick Dix San Marcos High School, Texas, james.p.dix@gmail.com Follow this and additional works at: Recommended Citation Dix, James Patrick (13) "Existence of the limit at infinity for a function that is integrable on the half line," Rose-Hulman Undergraduate Mathematics Journal: Vol. 14 : Iss. 1, Article 1. Available at:

2 Rose- Hulman Undergraduate Mathematics Journal Existence of the limit at infinity for a function that is integrable on the half line James Patrick Dix a Volume 14, No. 1, Spring 13 Sponsored by Rose-Hulman Institute of Technology Department of Mathematics Terre Haute, IN mathjournal@rose-hulman.edu a San Marcos High School, San Marcos, TX 78666, USA. james.p.dix@gmail.com

3 Rose-Hulman Undergraduate Mathematics Journal Volume 14, No. 1, Spring 13 Existence of the limit at infinity for a function that is integrable on the half line James Patrick Dix Abstract. It is well known that for a function that is integrable on [, ), its limit at infinity may not exist. First we illustrated this statement with an example. Then, we present conditions that guarantee the existence of the limit in the following two cases: When the integrable function is non-negative, if the first, second, third, or fourth, derivative is bounded in a neighborhood of each local maximum of f, then the limit exists. Without the non-negative constraint, if an integrable function has a bounded derivative on the entire interval [, ), then the limit exists. Acknowledgements: I would like to thank the anonymous referee for his/her valuable comments that helped me improve this article. Also I want to thank my teachers at the San Marcos High School and at the Honors Math Camp at Texas State University for providing me with the knowledge and discipline to approach these problems. In addition, I want to thank my family their support and encouragement.

4 Page RHIT Undergrad. Math. J., Vol. 14, No. 1 1 Introduction As a motivation for this article, we consider the rate of consumption of a non-renewable resource of which only a finite amount is available. Let f be consumption rate. Then the amount consumed of the resource from time zero to time t is represented by t f(s) ds. Since only a finite amount of the resource is available, we have the condition f(t) dt = A (A ± ), (1.1) which is the main assumption in this work. Does the consumption rate approach a constant (maybe zero) as t approaches infinity? Under what conditions does this limit exist? Another motivation for our research is considering a continuous probability distribution function defined on the interval [, ), and with finite mean. What is the long time behavior of the probability distribution function? The discrete version of (1.1) corresponds to a convergent series n= a n < ; in which case the terms a n must approach zero as n approaches infinity. However, assumption (1.1) does not guarantee the existence of lim f(t), as shown in the following example. 1 area=1/1 area=1/ area=1/3 Figure 1: The integral is finite, but lim f(t) does not exist Example 1.1. Let f(t) be the function plotted in Figure 1. Note that f(s) ds = k=1 1/k < [, Theorem 9.11]. Therefore, condition (1.1) is satisfied. However, the function f(t) oscillates without approaching any number as t approaches infinity; thus lim f(t) does not exist. Recall that a function has limit at infinity if and only if its limit superior and its limit inferior are equal [1, Theorem 6.3d]. These two limits are defined as follows: lim sup f(t) = lim sup{u(s) : s t}, lim inf f(t) = lim inf{u(s) : s t}. In this example, lim sup f(t) = 1 and lim inf f(t) =. Thus lim f(t) does not exist. We did not find any references for the existence of this limit; so made our goal to find conditions for the existence of lim f(t). Our main tools are standard calculus results. A trivial way to guarantee the existence of lim f(t) is by assuming that f is monotone. In this case f is of a constant sign on an interval [t, ). If f is positive, then for satisfying

5 RHIT Undergrad. Math. J., Vol. 14, No. 1 Page 3 (1.1), f must be non-increasing and lim f(t) =. If f is negative, then for satisfying (1.1), f must be non-decreasing and lim f(t) =. If f is identically zero, its limit is zero. Furthermore, monotonicity of f is implied by a higher-order derivative being of constant sign on an infinite interval. Assume that the nth derivative is of constant sign on an interval [t n, ). Then the (n 1)th derivative is monotone and of constant sign on some interval [t n 1, ). Then we show that the (n )th derivative is monotone and of constant sign on some interval [t n, ). By repeating this process, show that f is monotone on some interval [t, ). However, we want to find conditions weaker than the ones above. The existence of the limit of f is established for the following two cases: When f is non-negative, if the first, second, third, or forth derivative is bounded, by the same constant on neighborhoods of all points of relative maximum of f, then lim f(t) exists; see Theorems.3,.4,.5, and.7. When f is allowed to have negative values, if any derivative is bounded on [, ), then lim f(t) exists; see Theorem.9 and Corollary.11. To address probability distribution functions whose mean is finite, we present Corollary.1. Results The following lemmas will be used in proving our main results. Lemma.1. Let f be continuous on [, ), and lim inf f(t) lim sup f(t). Then there exists a constant α, satisfying lim inf f(t) < α < lim sup f(t), such that for every β >, we can define a sequence of relative maxima {f(t k )} such that f(t k ) > α, t k+1 > t k + β for k = 1,,.... Proof. Since the limit superior and the limit inferior are real numbers, not equal to each other, there exists a constant α such that lim inf f(t) < α < lim sup f(t). From the definition of limit superior, there exists a sequence of values of f(t) approaching the limit superior, and another sequence of values of f(t) approaching the limit inferior. Recursively, we define the infinite sequences {v j } and {w j } such that v 1 < w 1 < v < w < +, with f(v k ) > α and f(w k ) < α. Since f is continuous on each closed interval [w k, w k+1 ], by the extreme value theorem [, Theorem 3.1], f assumes its maximum at some value t k (w k, w k+1 ) and f(t k ) f(v k ) > α. Note that lim k t k =. Then we define a subsequence, also denoted by {t k }, such that t k+1 > t k + β k.

6 Page 4 RHIT Undergrad. Math. J., Vol. 14, No. 1 This completes the proof. Lemma.. Let f(t) be a continuous function for t, that has a local maximum at value t >. Then the following items hold: (1) If f (t ) is defined, then f (t ) = ; () If f (t ) is defined and f is continuous on an interval [t k, t k + λ) or (t k λ, t k ], with λ >, then f (t ) ; (3) If f (t ) is defined, then there is no information on the sign of f (t ); it can be positive, negative, or zero. Proof. Statement (1) follows from [, Theorem 3.]. Statement () is proven by contradiction. Suppose that f (t ) >. Then by continuity, f is positive on an interval containing t k ; therefore, f is concave up on this interval, which contradicts f(t k ) being a local maximum. Statement (3) is illustrated with examples: f 1 (t) = 1 3 t3 3 t + t has a local maximum at t = 1 and f (1) =. f (t) = 1 3 t3 + 3 t t has a local maximum at t = and f () =. f (t) = (t 1) has a local maximum at t = 1 and f () =. Theorem.3. Assume that f(t) is a continuous and non-negative function satisfying (1.1). Also assume that there exist positive constants M and λ such that at each local maximum f(t k ), f is continuous on [t k, t k + λ] and differentiable on (t k, t k + λ) satisfying f (t) M on (t k, t k + λ). Then lim f(t) exists. Proof. By contradiction assume that lim f(t) does not exist and show the integral in (1.1) approaches +. Since f, by Lemma.1, there exists α > and a sequence of local maxima {f(t k )} satisfying f(t k ) > α, t k+1 > t k + β, with β = min{λ, α/m}. At each t k, we draw a right triangle with vertices at (t k, ), (t k, α) and (t k, t k + β); see Figure. α t k t area=αβ/ k+1 Figure : Sequence of local maxima with triangles below Note that the definition of β does not allow the triangles to overlap, and each triangle has area αβ/. If the graph of f(t) stays above the hypotenuse of each triangle, then f(t) dt k=1 αβ =

7 RHIT Undergrad. Math. J., Vol. 14, No. 1 Page 5 which will complete the proof. Next we prove that the graph of f(t) stays above the hypotenuse. Note that each hypotenuse has slope with absolute value greater than of equal to M. Since f(t k ) > α, if there is a point t, in the base of the triangle, such that f(t) is below or on the hypotenuse, by the Mean Value Theorem there is a point s between t k and t, where f (s) > M. This contradicts the assumption that f (s) M; therefore f(t) stays above the hypothenuse. This completes the proof. Theorem.4. Assume that f(t) is a continuous and non-negative function satisfying (1.1). Also assume that there exist positive constants M and λ such that at each local maximum f(t k ), the functions f and f are continuous on [t k, t k +λ], and f is differentiable on (t k, t k + λ) satisfying f (t) M on (t k, t k + λ). Then lim f(t) exists. Proof. By contradiction assume that lim f(t) does not exist and show the integral in (1.1) approaches +. Since f(t), by Lemma.1, there exists α > and a sequence of local maxima {f(t k )} satisfying f(t k ) > α, t k+1 > t k + β, with β = min{λ, α/m}. α t k t area=αβ/3 k+1 Figure 3: Sequence of local maxima with parabolas Below each local maximum (t k, f(t k )), we draw the right branch of the parabola g(t) = α α(t t k ) /β, with vertex at (t k, α); see Figure 3. By integrating g(t), from t k to t k + β, we obtain that the area of the right lobe of the parabola has value αβ/3. The definition of β assures that the right lobes of the parabolas do not overlap. If the graph of f(t) stays above the parabola, then αβ f(t) dt =, 3 which will complete the proof. Next we prove that the graph of f(t) stays above the right lobe the parabola. Recall that f(t k ) > α, f (t k ) = (because f(t k ) is a local maximum), and f (t) M. By the Taylor theorem for f(t) at the point t k with t k < t < t k + λ, there is a point c, t k < c < t, such that k=1 f(t) = f(t k ) + f (t k )(t t k ) + f (c) (t t k) > α + M (t t k) α α(t t k) β = g(t), which is the parabola defined above. This completes the proof.

8 Page 6 RHIT Undergrad. Math. J., Vol. 14, No. 1 Theorem.5. Assume that f(t) is a continuous and non-negative function satisfying (1.1). Also assume that there exist positive constants M and λ such that at each local maximum f(t k ), the functions f, f and f are continuous on [t k, t k + λ], and f is differentiable on (t k, t k + λ) satisfying f (t) M on (t k, t k + λ). Then lim f(t) exists. Proof. By contradiction assume that lim f(t) does not exist and show that the integral in (1.1) approaches +. Since f(t), by Lemma.1, there exists α > and a sequence of local maxima {f(t k )} satisfying f(t k ) > α, t k+1 > t k + λ. Note that by Lemma., f (t k ) = and f (t k ). We consider the only two possible cases for the sequence {f (t k )} k=1. Case 1: The sequence {f (t k )} k=1 is bounded below by a constant γ. Then for t k < c < t k + λ, and using the bound M, we have f (c) = f (t k ) + c t k f (s) ds γ + M(c tk ) γ + λm. By the Taylor theorem for f at t k with t k < t < t k + λ, there is value c with t k < c < t such that f(t) = f(t k ) + f (t k )(t t k ) + f (c) (t t k) > α + (γ + λm) (t t k). When we restrict t to satisfy t t k < λ and (γ + λm)(t t k ) / α/, we have that f(t) > α/ on infinitely many disjoint intervals of the form [t k, t k + min{λ, α/(γ + M)}]. Then it follows that f(t) =. This completes the proof for case 1. Case : The sequence {f (t k )} k=1 is unbounded below. Then there are infinitely many k s such that f (t k ) < λm. Let [t k, t k ] be the largest interval containing t k such that f (t) on this interval. Claim: t k t k λ. On the contrary suppose that t k t k < λ, then f (t) M on [t k, t k ]. Recall that f (t k ) < λm 3 and f (t k ) =. An application of the mean value theorem yields a value c (t k, t k ) such that f (c) > M. This contradiction proves the claim. From this claim, the graph of f is concave down on the interval [t k, t k ]. Then the line that connects (t k, f(t k )) and (t k + λ, f(t k + λ)) lies under the graph of f, so the area under the graph of f and under the straight line satisfy tk +λ f(t) dt f(t k) + f(t k + λ) t k λ > + α λ. The second inequality holds because f(t k + λ) and f(t k ) > α.

9 RHIT Undergrad. Math. J., Vol. 14, No. 1 Page 7 Since t k + λ < t k+1, there are infinitely many disjoint intervals, of length λ, where the above estimate holds. It follows that f(t) =. This proves case and completes the proof. Remark.6. The assumptions for Theorems.3.5 are stated on right intervals [t k, t k + λ]. However, we can reach the same conclusion using left intervals [t k λ, t k ], or a combination of these two types of intervals. In the next theorem, we do not have this choice because the third derivative is required to be bounded on both sides of t k. Theorem.7. Assume that f(t) is a continuous and non-negative function satisfying (1.1). Also assume that there exist positive constants M and λ such that at each local maximum f(t k ), the functions f, f, f and f are continuous on [t k λ, t k +λ], and f is differentiable on (t k λ, t k + λ) satisfying f (4) (t) M on (t k λ, t k + λ). Then lim f(t) exists. Proof. By contradiction assume that lim f(t) does not exist and show that the integral in (1.1) approaches +. By Lemma.1, there exists α > and a sequence of local maxima {f(t k )} satisfying f(t k ) > α, t k+1 > t k + λ. By Lemma., for all k: f (t k ) = and f (t k ). Let [t k, t k ] be the largest interval containing t k such that f (t) on this interval. We consider the two possible cases for the lengths of these intervals. Case 1: lim k t k t k =. From the definition of limit, there exists an index k such that t k t k < λ for all k k ; thus [t k, t k ] [t k λ, t k + λ]. Claim: lim k f (t k ) =. Note that f (t k ) = f (t k ) = and f is continuous on [t k, t k ]. By the extreme value theorem, f attains its minimum at some point c k (t k, t k ). Then f (c k ) f (t k ). Using that f (c k ) = and the bound M, for t [t k λ, t k + λ], we have f (t) = f (c k ) + t Thus the graph of f intersects the t-axis at t k By the fundamental theorem of calculus, f (t k ) f (c k ) = t k c k c k f (4) (s) ds M t c k λm. f (t) dt and at t k, and has slope bounded by λm. t k c k f (t) dt λm(t k c k ), and similarly, Therefore, f (c k ) f (t k) λm(c k t k). λm(t k c k ) f (t k ) f (c k ) λm(t k c k ), λm(c k t k) f (c k ) f (t k) λm(c k t k).

10 Page 8 RHIT Undergrad. Math. J., Vol. 14, No. 1 Since f (t k ) = f (t k ) =, it follows that f (c k ) λm(t k t k ). Then the inequality f (c k ) f (t k ) implies lim k f (t k ) =, which proves the claim. From this claim, for each β >, there exists an index k such that f (t k ) β for all k k. By the Taylor theorem, for f at t k with t k λ < t < t k + λ, we have f(t) = f(t k ) + f (t k )(t t k ) + f (t k ) (t t k) > α + β t t k λm t t k f (c) (t t k) 3 6 By restricting t to satisfy t t k < λ, β t t k / α/3 and λm t t k 3 /6 α/3, we have that f(t) > α/3 on infinitely many intervals, each one of length min{λ, α/(3β), 3 3α/(λM)}. Since these intervals are disjoint, it follows that f(t) =. This completes the proof for case 1. Case : There exists a positive constant γ such that t k t k γ for infinitely many k s. Then the graph of f is concave down on [t k, t k ]. Since f(t) and f(t k) > α, the area under the graph can be estimated as tk t k f(t) dt + t k t k f(t) dt f(t k ) + f(t k) (t k t k) + f(t > + α (t k t k) αγ. k ) + f(t k) (t k t k ) Since intervals [t k, t k ] cover a set of intervals whose lengths add up to infinity, f(t) =. This proves case. It also completes the proof. Remark.8. The above methods for proving Theorems.3.7 does not seem to work when only the fifth derivative is bounded. However, we are unable to find an example where the fifth derivative is bounded, f(t), and lim f(t) does not exist. Next we consider a function which may have positive and negative values, however we need to assume differentiability on the entire interval [, ). Theorem.9. Assume that f(t) is a differentiable function satisfying (1.1), and that there exists a positive constant M such that f (t) M on for all t. Then lim f(t) =. Proof. To prove that lim f(t) =, we show that lim sup f(t) =. By contradiction, assume that lim sup f(t) = α for some positive constant α. Then there exists a sequence f(t k ) approaching either α or α, and {t k } as k. First assume that lim k f(t k ) = α. As in Lemma.1, we select a subsequence such that t k+1 t k +(α/(3m)) and f(t k ) α/3, k = 1,,....

11 RHIT Undergrad. Math. J., Vol. 14, No. 1 Page 9 The mean value theorem and the assumption f (s) M guarantee the existence of an interval I 1 with center t 1 and radius α/(3m) where f(t) α/3. Similarly, at t, we obtain an interval I with center t and radius α/(3m) where f(t) α/3. The same process yields a sequence of non-overlapping intervals I 1, I, I 3,..., each one of length α/(3m). Let I k = [a k, b k ]; then a k and b k approach infinity as k approaches infinity. Since f(t) α/3 on [a k, b k ], we have bk f(t) dt bk = f(t) dt α a k a k 9M. However, from (1.1), for α /(9M) > there exists a value T such that a k, b k T implies bk f(t) dt ak f(t) dt = bk a k f(t) dt < α 9M. This contradiction implies α =. Now assume that lim k f(t k ) = α Note that lim k ( f(t k )) = α, and that the function f satisfies (1.1) with A instead of A. Using the first part of this proof, for f, we conclude that α =. This completes the proof. Theorem.1. Assume that f(t) satisfies (1.1), and has continuous derivatives on [, ). If the k-th derivative satisfies sup t f (k) (t) =, then sup t f (p) (t) = for all p k. Proof. On the contrary suppose that for some integer n, sup t f (n) (t) =, and that for all t there exists a constant M such that f (n+1) (t) M. Then for the value α = M ( n + n ) there exists a t 1 such that f (n) (t 1 ) α + 1. Using that the absolute value of the slope of f (n) is bounded by M, by the mean value theorem, there exists an interval I n, with center t 1 with radius α/m such that f (n) (t) 1 for all t I n. On I n, the function f (n 1) is strictly monotonic because the absolute value of its derivative is greater than or equal to 1. Let Ĩ n = {t I n : f (n 1) (t) < 1}. This set can be empty, or an interval of length at most, because the absolute value of the slope of f (n 1) is at least 1. If the set I n \ Ĩn consists of a single subinterval of I n, we let I n 1 = I n \ Ĩn. (In set notation A \ B consists of all the elements that are in A but not in B). If the set I n \ Ĩn consists of two subintervals, we let I n 1 be larger of the two subintervals. In both cases I n 1 I n, and I n 1 is an interval of length at least 1 ( α M ). On I n 1, the function f (n ) is strictly monotonic and the absolute value of its slope is greater than or equal to 1. Let Ĩn 1 = {t I n 1 : f (n ) (t) < 1}. Repeating the above process, we define an interval I n I n 1 whose length is at least 1 ( 1 ( α M ) ).

12 Page 1 RHIT Undergrad. Math. J., Vol. 14, No. 1 Repeating this process, n times, we obtain a interval I whose length is at least, such that f(t) 1 for all t I. Let [a 1, b 1 ] = I. Since f (n) is continuous and unbounded on [, ), for the same α defined above, there exists a t t 1 + (α/m) such that f (n) (t ) α + 1. As above we obtain an interval [a, b ] of length at least, such that f(t) 1 for all t [a, b ]. Repeating the above process we obtain a sequence of non-overlapping intervals [a k, b k ], each one of length at least. Since f(t) 1 and is continuous on [a k, b k ], we have bk f(t) dt = bk f(t) dt. a k a k However, from (1.1), for the positive number there exists a value T such that a k, b k T implies bk ak f(t) dt f(t) dt bk = f(t) dt <. a k This contradiction implies that f (n+1) can not be bounded. The proof is complete. As a consequence of Theorems.9 and.1, we have the following result for functions that may have negative values. Corollary.11. Assume that f(t) satisfies (1.1), and has continuous derivatives. If the k-th derivative satisfies sup t f (n) (t) <, then sup t f (n) (t) < for all k n. Furthermore, lim f(t) =. We conclude this section by considering a continuous probability distribution function defined on the interval [, ), and having finite mean. Corollary.1. Assume that the probability distribution function g is defined on the interval [, ) and its mean satisfies tg(t) dt = µ <. If for any pair of derivatives g (n), g (n+1), we have sup t g (n) (t) + tg (n+1) (t) <, then lim tg(t) =. Proof. By setting f(t) = tg(t) in Corollary.11, it follows that if for some n, sup t g (n) (t)+ tg (n+1) (t) <. Then lim tg(t) =. References [1] Watson Fulks; Advanced calculus: An introduction to analysis, John Wiley, New York - London, 1961.

13 RHIT Undergrad. Math. J., Vol. 14, No. 1 Page 11 [] Ron Larson, Robert P. Holster, Bruce H. Edwards; Calculus with analytic geometry, Houghton Mifflin Co., Boston - New York, 6. [3] Maurice D. Weir, Joel Hass, Frank R. Giordano; Thomas Calculus, eleventh edition, Pearson, Boston San Francisco, 5.