Evaluate limit of this integral using positive summability kernels

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SUMMARY

The integral evaluation using positive summability kernels reveals that the expression simplifies to ##-f'(0)## when applying integration by parts. The antiderivative of ##\varphi'(nx)## is confirmed as ##\frac1{n}\varphi(nx)##, leading to the conclusion that ##n\varphi(nx)## serves as a positive summability kernel. The discussion highlights the importance of verifying the kernel's properties, particularly when substituting functions like ##f(x)=x^r## for further validation.

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Homework Statement
Let ##\varphi## be defined by ##\varphi(x)=\frac{15}{16}(x^2-1)^2## for ##|x|<1## and ##\varphi(x)=0## otherwise. Let ##f## be a function with a continuous derivative. Find the limit $$\lim_{n\to\infty}\int_{-1}^1n^2\varphi '(nx)f(x)dx.$$
Relevant Equations
Positive summability kernels, see e.g. Wikipedia.
Integrating the integral by parts, using that the antiderivative of ##\varphi'(nx)## is ##\frac1{n}\varphi(nx)##, I get
$$\big[n\varphi(xn)f(x)\big]_{-1}^1-\int_{-1}^1 n\varphi(nx)f'(x)dx=0-\int_{-1}^1 n\varphi(nx)f'(x)dx.$$ I used the fact that ##\varphi(n)## and ##\varphi(-n)## both equal ##0##, since ##n\geq 1##.

However, I'm stuck here. This is a problem from a section on positive summability kernels, but I have been unable to verify what the kernel is in this exercise, if there is any. Appreciate any help.
 
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I solved it I think. ## n\varphi(nx)## is a positive summability kernel and the integral therefor evaluates to ##-f'(0)##.
 
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You could try checking by plugging in ##f(x)=x^r##.
 
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