Prove or give a counterexample: If f is differentiable on a neighborhood of x0,
ID: 1892968 • Letter: P
Question
Prove or give a counterexample: If f is differentiable on a neighborhood of x0, then f satisfies a Lipshitz condition on some neighborhood of x0Lots of details please.
Explanation / Answer
----------------->Let f:R -> R be a function where f(x + y) = f(x) * f(y). Let f be differentiable at zero and f is not identically zero. Then f ' (0) exists. ----------------->Then f ' (0) = Lim (h->0) [f(h) - f(0)] / h. ----------------->Claim: f(0) = 1 --------------->Proof of claim: f(0) = f(0 + 0) = f(0) * f(0) so divide both sides by f(0). Then f(0) = 1. ----------------->Then f ' (0) = Lim (h->0) [f(h) - 1] / h --------------->First need to show that f ' (x) exists for all x. By definition we need to show that following limit exists. ---------------->f ' (x) = Lim (h->0) [f(x+h) - f(x)] / h ------------------>f ' (x) = Lim (h->0) [f(x)f(h) - f(x)] / h **** By the property f(x + y) = f(x) * f(y) ------------->f ' (x) = Lim (h->0) [f(x) (f(h) - 1)] / h ---------------->f ' (x) = f(x) * Lim (h->0) [f(h) - 1] / h ------------->f ' (x) = f(x) * f ' (0) -------------->By assumption, f(x) exists for all x and f ' (0) exists since f is differentiable at 0. Therefore, f ' (x) exists for all x. Thus, f is differentiable everywhere and f ' (x) = f(x) * f ' (0). --------------->Now need to show f(x) = e^(cx) where c = f ' (0). ------------->Start with f ' (x) = f(x) * c. In calculus, this looks like a separable differential equation. Let's use the following notation: f ' (x) = dy/dx and f(x) = y. ---------------->dy/dx = yc ---------------->dy/dx * 1/y = c --------------->INT {1/y} dy = INT {c} dx -------------->ln y = cx ---------------->e^(ln y) = e^(cx) ------------>y = e^(cx) -------------->Now change y back to f(x) ------------->f(x) = e^(cx)
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