WebOct 26, 2024 · In this improvised video, I show that if is a function such that f (x+y) = f (x)f (y) and f' (0) exists, then f must either be e^ (cx) or the zero function. It's amazing how we can derive all that ...
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WebSep 26, 2010 · Let f be a real-valued function on R satisfying f (x+y)=f (x)+f (y) for all x,y in R. If f is continuous at some p in R, prove that f is continuous at every point of R. Proof: Suppose f (x) is continuous at p in R. Let p in R and e>0. Since f (x) is continuous at p … WebViewed 3k times 2 Consider the function f: R 2 → R given by f ( x, y) = max ( x, y). (That is, f ( x, y) is the larger of x and y, so f ( − 3, 2) = 2, f ( 1, 4) = 4, and f ( − 3, − 2) = − 2 .) (assume that R 2 has sup metric) prove that f is continuous.
WebJan 4, 2015 · A function f : X → Y is continuous if, for every x ∈ X and every open set U containing f (x), there exists a neighborhood V of x such that f (V) ⊂ U. Proof: Let C be a closed subset of Y, s.t, C ⊂ Y. Clearly, if C is closed, the set Y-C is open since the compliment of a closed set is an open set (Theorem 6.5). WebFeb 21, 2024 · Homework Statement Let f be a continuous function lR (all real numbers) --> lR such that f (x+y) = f (x) + f (y) for x, y in lR. prove that f (n) = n*f (1) for all n in lN (all natural numbers) Homework Equations f is continuous also note and prove that f (0) = 0 The Attempt at a Solution Edit:
WebSo is continuous. (inverse image of closed is closed). This direction only uses compactness of . For the other direction we only need the Hausdorffness of : The diagonal is closed iff … WebFeb 7, 2024 · Overview of basic facts about Cauchy functional equation (1 answer) Show if f ( x + y) = f ( x) + f ( y) for all x, y and f is Lebesgue measurable, then f is continuous. …
WebSo now we see that if ( x n, y n) ∈ G ( f), ( x n, y n) → ( x, y), then y n → f ( x n) as defined by G ( f) and x n → x, f ( x n) → y. Since f is assumed to be continuous, f ( x n) → f ( x) so y = f ( x). Therefore ( x, y) ∈ G ( f) and we conclude G ( f) is closed. Share Cite Follow edited Feb 22, 2016 at 22:06 YoTengoUnLCD 13.1k 4 39 99
WebOct 5, 2024 · Let f ( x, y) be a continuous real-valued function on the unit square [ 0, 1] × [ 0, 1]. Show that h ( x) = max { f ( x, y): y ∈ [ 0, 1] }, is also continuous. Answer. Since f ( x, y) is continuous, then max { f ( x, y) } is also continuous on [ 0, 1] × [ 0, 1]. Thus for any fixed values of y ∈ [ 0, 1] , max { f ( x, y) } is also continuous . fawn mortality rate for whitetail deerWebApr 11, 2011 · The question states: Give two different examples of f:R->R such that f is continuous and satisfies f(x+y)=f(x)+f(y) for every x,y e R. Find all continuous functions f:R->R having this property. Justify your answer with a … friendly letter writing worksheetWebYou have already shown: if f (x+ f (y))= f (x+ y)+1 and if f is surjective, then f (z) = z + 1 for all z. Now it remains to show that the function given by f (x) = x+1 , is injective and … friendly lifeWebAug 16, 2024 · So f must be a linear function. The only linear functions Q → Q are of the form f ( x) = a x. For such a function, we must have f ( x + f ( y)) = f ( x) + y; that is, we must have a ( x + a y) = a x + y. So we must have a 2 = 1. So the only functions that could possibly work are f ( x) = x and f ( x) = − x. friendly library near meWebLet f ( x, y) = x y ( x 2 − y 2) / ( x 2 + y 2) with f ( 0, 0) = 0 . The question was asking to proof f ( x, y) continuous everywhere. One way to solve it was to just change x = r cos ( θ), y = r sin ( θ) and solved it. First question: However, is there is a way to just solve it through x, y without the transformation of coordinates? friendly license s.lWeb2 Answers Sorted by: 4 The function f ( x, y) = x is continuous since given ϵ > 0 and ( a, b) ∈ R 2, setting δ = ϵ makes f ( x, y) − f ( a, b) = x − a = ( x − a) 2 ≤ ( x − a) 2 + ( y − … friendly letter writing templatesWebMay 23, 2015 · The solution I have is that f is not continuous in . (The solution doesn't say more than that.) However, the result I got is that is continuous in . Here's my approach: … friendly lgbt countries