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Thread: integral of a sequence of functions

  1. #1
    Super Member Anonymous1's Avatar
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    integral of a sequence of functions

    Prove that if on an interval $\displaystyle [a,b]$ the functions $\displaystyle g_{n}(x)$ are continuous and converge uniformly to $\displaystyle g(x),$ then on $\displaystyle [a,b]$

    $\displaystyle \int_a^x g_{n}(t) dt$ converges uniformly to $\displaystyle \int_a^x g(t) dt.$
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    Member Abu-Khalil's Avatar
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    $\displaystyle g(x)$ is continuous and therefore integrable over $\displaystyle [a,b]$ 'cause $\displaystyle g_n\to g$ uniformly and each $\displaystyle g_n$ is continuous there.

    Let $\displaystyle f_n(x)=\int_a^xg_n(t)dt,f(x)=\int_a^xg(t)dt$. So we wanna prove that $\displaystyle f_n\to f$ uniformly.

    Let $\displaystyle \epsilon>0$, then $\displaystyle \exists N:\forall n\geq N,\left|g_n(x)-g(x)\right|<\epsilon,\forall x\in[a,b]$.
    Notice that $\displaystyle \forall x\in[a,b]$ we have $\displaystyle \left|f(x)-f_n(x)\right|=\left|\int_a^xg_n(t)dt-\int_a^xg(t)dt\right|\leq \int_a^x\left|g_n(t)-g(t)\right|dt\leq \int_a^x\epsilon dt\leq \epsilon(b-a).$
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    Quote Originally Posted by Anonymous1 View Post
    Prove that if on an interval $\displaystyle [a,b]$ the functions $\displaystyle g_{n}(x)$ are continuous and converge uniformly to $\displaystyle g(x),$ then on $\displaystyle [a,b]$

    $\displaystyle \int_a^x g_{n}(t) dt$ converges uniformly to $\displaystyle \int_a^x g(t) dt.$



    $\displaystyle \forall\,\epsilon>0\,\,\exists\,N_{\epsilon}\!\!\i n\mathbb{N}\,\,s.t.\,\,\left|g_n(x)-g(x)\right|<\frac{\epsilon}{b-a}\,\,\forall\,n>N_{\epsilon}\mbox{ and }\forall\,x\in[a,b]$ , so :

    $\displaystyle \left|\int\limits_a^x g_{n}(t) dt-\int\limits_a^x g(t) dt\right|\leq \int\limits_a^b \left|g_n(t)-g(t)\right|dt<\int\limits_a^b\frac{\epsilon}{b-a} dt=\epsilon\,,\,\,\forall\,n>N_{\epsilon}$ $\displaystyle \mbox{ and }\,\forall\,x\in[a,b]$.

    Tonio
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