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Question: Could someone give an example of a sequence of uniformly continuous real-valued functions on the reals such that they converge point-wise to a function that is continuous but not uniform continuous.

My attempt so far: I managed to prove this is true in the case of uniform convergence, so I'm convinced there is an example. I considered triangles such that they were symmetrical on the y-axis such that they got taller and closer however this converges point-wise to 0 which is uniform continuous

WhizKid
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I'm not convinced this is the simplest example (I certainly wouldn't want to make it explicit), but it was fun :)

Lines are uniformly continuous and quadratics are not. Also, continuous implies is uniformly continuous on a compact domain. In addition, it's not too hard to show that if you glue two uniformly continuous functions together, the result is uniformly continuous.

Now create a function that approximates a quadratic by being identical near the vertex and at some point on each side breaking off into the tangent lines. Therefore, considering a sequence of such functions whose "breakoff points" have $x$-values that go to $\pm\infty$; these satisfy the premises and get the conclusion, so you're good.

Eric Stucky
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  • btw what do you mean by lines are uniformly continuous? Im guessing your referring to something like x^2 which isnt uniform continuous, now if I can created a sequence of functions that are tangents to points on it and the as I go further in the sequence, I get like a "zig-zag" of continuous lines on closed bounded intervals mimicking the shape of x^2, Im guessing this converges to x^2 pointwise due to it being a linear approximation at those certian points? Is this the idea? Thanks – WhizKid Oct 30 '13 at 01:04
  • @WhizKid: Ah, sorry. I meant that each function in the sequence has just two $x$-values (one per side) at which the function stops being quadratic and starts being a line; this might as well be the tangent line at that point. I never said it was a good approximation :P – Eric Stucky Oct 30 '13 at 01:06
  • im unable to get what you mean visually, it would be great if you could show me it show how as I'm sure once I've got the shape I can construct a function. – WhizKid Oct 30 '13 at 01:11
  • @WhizKid: Does this help? http://www.wolframalpha.com/input/?i=plot+Piecewise%5B%7B+%7B-2x-1%2Cx%3C-1%7D%2C%7Bx%5E2%2C-1%3C%3Dx%3C%3D1%7D%2C%7B2x-1%2Cx%3E1%7D+%7D%5D+ – Eric Stucky Oct 30 '13 at 01:16
  • right so the next piecewise addition will be -4x-4 and 4x-4, so its like a good linear approximation :p ?? – WhizKid Oct 30 '13 at 01:23
  • Hm, I'm not sure if I know what you're asking. Every function in the sequence I've created has only three pieces, it's just that the break points (in the link these are $\pm 1$) get farther and farther from the vertex. If that's what you meant, then yes. But I wasn't adding more linear pieces at each step; that would probably work if done right but it would be even messier. – Eric Stucky Oct 30 '13 at 01:27
  • ok, I totally understand what you did, thanks! thats very nice :) – WhizKid Oct 30 '13 at 01:31