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Problem: Let $ f: [0,\infty) \to \mathbb{R} $ be twice continuously differentiable and suppose that $ \int_0^\infty f(x)^2 dx $ , $ \int_0^\infty f''(x)^2 dx $ converge. Then $ \int_0^\infty f'(x)^2 dx $ converges

Attempt: $ \int_{0}^{ k} f'(x)^2 dx = \{ u = f' , u = f, v = f' , v' = f'' \} = f'(x) \cdot f''(x) |_0^k - \int_0^k f\cdot f'' $ [ missing more arguments ].

How can I continue from where I've stopped? It seems to me I need to use some kind of an inequality. If it is Cauchy-Schwartz then I haven't learned about Cauchy-Schwartz Inequality yet, so is there another way?

Edit: Apparently there's the same question here with answers which use Cauchy-Schwartz Inequality: If $f$ and $f''$ are square integrable is $f'$? But my question is whether it is possible to prove the same theorem without using that inequality? as I haven't learned about it yet in relation to integrals.

hazelnut_116
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  • Here is the same question only with $\cal L^1$ integrability: https://math.stackexchange.com/q/3867307/42969. I assume that you can proceed similarly. – Martin R Nov 10 '21 at 12:13
  • Is $ f''$ continious? – Botnakov N. Nov 10 '21 at 12:20
  • @MartinR I haven't learned about Lebesgue integrability ( and looking at the details of it, there are more things I haven't learned like Fubini-Tonelli theorem ) but In case there'll be no answer, I'll try to follow that question's answer. – hazelnut_116 Nov 10 '21 at 12:25
  • @BotnakovN. You are correct, there was an error in the original question, now $ f $ is twice continuously differentiable – hazelnut_116 Nov 10 '21 at 12:30
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    Apparently there is a simpler answer: https://math.stackexchange.com/q/633332/42969 (and you were on the right track). – Martin R Nov 10 '21 at 12:31
  • @MartinR How can I avoid using Cauchy-Schwartz Inequality? I haven't learned about it in relation to integrals. – hazelnut_116 Nov 10 '21 at 12:32
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    You can use the inequality between geometric and arithmetic mean, see the comment here: https://math.stackexchange.com/questions/633332/if-f-and-f-are-square-integrable-is-f?noredirect=1#comment1334987_633357. – Martin R Nov 10 '21 at 12:34
  • Ok thanks, I'll try. I guess my question's a duplicate but I did try searching it before answering ( I didn't shorten my format when searching and probably that's why I haven't seen it ) – hazelnut_116 Nov 10 '21 at 12:37

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