Emmy Knox

2022-06-26

Operator on polynomials, antiderivative
We are given a linear map:
and two norms on $\mathbb{R}\left[X\right]$ : $||p|{|}_{\mathrm{\infty }}=\underset{t\in \left[0,1\right]}{sup}|p\left(t\right)|$, $||p|{|}_{1}={\int }_{0}^{1}|p\left(t\right)|dt$.
I want to check whether the map is continuous in these norms.
When it comes to the first norm, I get:
$||q|{|}_{\mathrm{\infty }}=\underset{t\in \left[0,1\right]}{sup}|q\left(t\right)|$
By Lagrange mean value theorem we have that there exists $a\in \left[0,1\right]$ such that ${q}^{\prime }\left(a\right)=\frac{q\left(t\right)-q\left(0\right)}{t}$, so $q\left(t\right)\le \underset{\left[0,1\right]}{sup}|{q}^{\prime }\left(t\right)|$.
And $||p|{|}_{\mathrm{\infty }}=\underset{t\in \left[0,1\right]}{sup}|p\left(t\right)|=\underset{t\in \left[0,1\right]}{sup}|{q}^{\prime }\left(t\right)|\ge ||q||$
Is that correct so far?

Punktatsp

Expert

Step 1
Your work is correct, but in my opinion this makes things more clear:
$q\left(x\right)={\int }_{0}^{x}p\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt.$
Then, for the $‖\phantom{\rule{thinmathspace}{0ex}}{‖}_{\mathrm{\infty }}$ norm
$|q\left(x\right)|\le {\int }_{0}^{x}|p\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt\le x\phantom{\rule{thinmathspace}{0ex}}‖p{‖}_{\mathrm{\infty }}\le ‖p{‖}_{\mathrm{\infty }}.$
Step 2
Fot the $‖\phantom{\rule{thinmathspace}{0ex}}{‖}_{1}$ norm ${\int }_{0}^{1}|q\left(x\right)|\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{1}|{\int }_{0}^{x}p\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt|\phantom{\rule{thinmathspace}{0ex}}dx\le {\int }_{0}^{1}{\int }_{0}^{1}|p\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt\phantom{\rule{thinmathspace}{0ex}}dx=‖p{‖}_{1}.$

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