- #1
SiddharthM
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I have several questions
1. Find an example of two sequences of functions, f_n and g_n such that they both converge uniformly to f, g on some set E but such that f_n * g_n does not converge uniformly on E.
Let f_n = x^2 for all n. and g_n=sinx/(xn). Since sinx/x is bounded by 1 it is easy to show that g_n converges uniformly to the zero function. however, the product is (xsinx)/n. This new function converges pointwise to the zero function. But, for any fixed n we can make (xsinx)/n as large as we please. So f_n * g_n is not uniformly convergent to the zero function on the real line. But f_n and g_n DO converge uniformly on the real line.
Is this correct?
2. consider f(x)=infinite sum from n=1 to infinity of {1/(1+(n^2)x)}
By the comparison test it is easy to show that the series converges absolutely for any positive nonzero x. For x=-1/(n^2) for any n, it's clear that f(x) doesn't converge. Otherwise the comparison test can be used to show that f(x) converges absolutely.
On what interval(s) does the series converge to f uniformly?
I put E= [c,infinity) with c>0. Then for all x in E we see that (n^2)x+1>(n^2)c+1>(n^2)c so that 1/((n^2)c)>1/((n^2)c+1)>1/((n^2)x+1). By convergence of p-series (p=2) and the wieirstrass m-test we see that the series converges to f uniformly on E. I take it that although c was arbitrarily chosen it does NOT follow that the series converges uniformly on (0,infinity)? This because f(x) is unbounded on (0,infinity) (b/c it blows up as it approaches 0) and the partial sums of the series are bounded on the nonzero positive reals. Since uniform convergence on the nonzero positive reals of a sequence (of partial sums of functions) of bounded functions implies a bounded limit function on the nonzero positive reals which is not the case we have here. So the convergence cannot be uniform on all of the positive numbers.
is this correct?
Please DO NOT tell me any answers, i ONLY want to know if the arguments I have given are correct. Thanks for your time.
1. Find an example of two sequences of functions, f_n and g_n such that they both converge uniformly to f, g on some set E but such that f_n * g_n does not converge uniformly on E.
Let f_n = x^2 for all n. and g_n=sinx/(xn). Since sinx/x is bounded by 1 it is easy to show that g_n converges uniformly to the zero function. however, the product is (xsinx)/n. This new function converges pointwise to the zero function. But, for any fixed n we can make (xsinx)/n as large as we please. So f_n * g_n is not uniformly convergent to the zero function on the real line. But f_n and g_n DO converge uniformly on the real line.
Is this correct?
2. consider f(x)=infinite sum from n=1 to infinity of {1/(1+(n^2)x)}
By the comparison test it is easy to show that the series converges absolutely for any positive nonzero x. For x=-1/(n^2) for any n, it's clear that f(x) doesn't converge. Otherwise the comparison test can be used to show that f(x) converges absolutely.
On what interval(s) does the series converge to f uniformly?
I put E= [c,infinity) with c>0. Then for all x in E we see that (n^2)x+1>(n^2)c+1>(n^2)c so that 1/((n^2)c)>1/((n^2)c+1)>1/((n^2)x+1). By convergence of p-series (p=2) and the wieirstrass m-test we see that the series converges to f uniformly on E. I take it that although c was arbitrarily chosen it does NOT follow that the series converges uniformly on (0,infinity)? This because f(x) is unbounded on (0,infinity) (b/c it blows up as it approaches 0) and the partial sums of the series are bounded on the nonzero positive reals. Since uniform convergence on the nonzero positive reals of a sequence (of partial sums of functions) of bounded functions implies a bounded limit function on the nonzero positive reals which is not the case we have here. So the convergence cannot be uniform on all of the positive numbers.
is this correct?
Please DO NOT tell me any answers, i ONLY want to know if the arguments I have given are correct. Thanks for your time.