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When $C[a,b]$ is the space of all real-valued continuous function defined on $[a,b]$ endowed with the uniform norm, for $F: C[a,b] \to C[a,b]$ given by $F(x)(t) = \int_a^t x(s)ds$ to show continuity can I argue that for all $t$ it holds that $$\lim_{n \to \infty} F(x_n)(t) = \lim_{n \to \infty} \int_a^t x_n(s)ds = \int_a^t x(s)ds = F(x)(t)$$ when $x_n$ converges to $x$ uniformly?

student24
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  • You'll need more detail in the middle equality; either expand on it further, or cite a result. Consider, for example, why it works when $x_n \to x$ uniformly. It doesn't work when convergence is pointwise! – Theo Bendit Jul 21 '19 at 16:03

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This argument is not sufficient to show that $F$ is continuous since you've only shown that $Fx_n \to Fx$ pointwise but you need uniform convergence of $Fx_n$ to $Fx$. You could try to add some details to the above argument to conclude that the convergence is in fact uniform in $t$.

However, I feel I should point out that for linear operators on normed spaces, it is often much easier to check that the operator is bounded to see that it is continuous. This means you would check that $$\|Fx\|_\infty \leq C \|x\|_\infty$$ for some $C$. This is easier since $$|Fx(t)| = \big| \int_a^t x(s) ds \big | \leq \int_a^t \|x\|_\infty ds \leq (b-a) \|x\|_\infty.$$

Rhys Steele
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  • Thanks for your help. Can you share what details one could add to conclude uniform convergence from pointwise one? – student24 Jul 21 '19 at 16:24
  • You wouldn't conclude it from the pointwise one. You would go through the argument you made to get pointwise convergence and check that the convergence is uniform. That is, you want to check that the way you pass the limit through the integral is uniform in $t$. This would probably look something like the series of inequalities in the other answer. – Rhys Steele Jul 21 '19 at 20:01
  • I see, I thought you meant there was a case where pointwise implies uniform convergence. Thanks for clarifying. – student24 Jul 21 '19 at 20:11
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You can prove it using the inequalities:

$\vert F(x_n)(t)-F(x)(t)\vert =\vert \int_a^t (x_n(s)-x(s))\vert \leq \int_a^t \vert x_n(s)-x(s) \vert ds \leq (b-a)\cdot \Vert x_n-x\Vert_\infty$

From which you'll obtain not only point-wise convergence (which is what you asked about), but also uniform convergence of $F(x_n)$ to $F(x)$, which is the metric on $C[a,b]$.

Keen-ameteur
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