Show that the norm of the maximum induces a Banach space

Question

Let $\mathbf{P}$ The set of polynomials of degree $\leq$ two, with real coefficients in $[0, 1]$, if $l=ax^2+bx+c$ in $\mathbf{P}$, we define $||l||_{\infty}=\max_{x\in[0, 1]}{\{|l(x)|\}}$, i want to show thtat $(\mathbf{P}, ||l||_{\infty})$ is a Banach space That is to say that there exists $q(x)\in \mathbf{P}$ s.t

$$\lim\limits _{n \to \infty} ||l_{n}-q(x)||_{\infty}=0$$

Where $\{l_{n}\}$ is Cauchy sequence. First we can see what the value of the limit can be and if this value belongs to the space.

Note that $l_{n} \in \mathbf{P}$, for all $n \in \mathbb{N}$, so $l_{n}=a_{n}x^2+b_{n}x+c_{n}$ with $a_{n}, b_{n}, c_{n} \in \mathbb{R}$, as $\{l_{n}\}$ is a Cauchy sequence, then for all $\epsilon>0$, there is $N\in \mathbb{N}$ such that if $n, m \geq N$ then

$$||l_{n}-l_{m}||_{\infty} < \epsilon$$

This means

$$|l_{n}-l_{m}|\leq\max_{x\in[0, 1]}{\{|l_{n}-l_{m}|\}} < \epsilon$$

So $$|l_{n}-l_{m}|< \epsilon$$

Then, we have that $\{l_{n}\}$ is a Cauchy sequence in $\mathbb{R}$ and as $(\mathbb{R}, |.|)$ is a Banach space then there exists $\lambda \in \mathbb{R}$ such that

$$\lim \limits _{n\to \infty} l_{n}=\lambda$$

So I think we can take $q(x)=0x^{2}+0x+\lambda$ (because $a, b \in [0, 1]$).

Finally, it only remains to verify that the initial limit is zero, so let $\epsilon>0$, we have that $\lim \limits _{n\to \infty} l_{n}=\lambda$, so there exists $N_{1}\in \mathbb{N}$ s.t

$$|l_{n}-q(x)|<\epsilon$$

So, how can I conclude the result?

Any help is greatly appreciated!

Answer 1

A short answer could be @geetha290krm comment, but let us prefer (and correct) your "naked hands" approach.

First, it is not true that "$\{l_n\}$ is a Cauchy sequence in $\mathbb{R}$ and [...] there exists $\lambda \in \mathbb{R}$ such that $\lim \limits _{n\to \infty} l_n=\lambda$", because each $l_n$ is a polynomial, not a real number.

However, this reasoning can be applied to $\{l_n(0)\}$ instead of $\{l_n\},$ thereby showing that $$\lim \limits _{n\to \infty} c_n=c\in\Bbb R.$$

Similarly, considering $\{l_n(1)\}$ and $\{l_n(1/2)\},$ you can derive $$\lim \limits _{n\to \infty} a_n+b_n+c_n=d\in\Bbb R$$ $$\lim \limits _{n\to \infty} a_n/4+b_n/2+c_n=e\in\Bbb R$$ and therefore, $$\lim \limits _{n\to \infty}b_n=b:=-3c-d+4e,$$ $$\lim \limits _{n\to \infty} a_n=a:=2c+2d-4e.$$ Finally, the polynomial $$l:=ax^2+bx+c$$ is the limit, for your norm $\|~\|_\infty,$ of $\{l_n\},$ because for every $n\in\Bbb N$ and $x\in[0,1],$ $$|(a_n-a)x^2+(b_n-b)x+(c_n-c)|\le k_n,$$ where $$k_n:=|a_n-a|+|b_n-b|+|c_n-c|\to0.$$

Answer 2

I think that some modifications on your notation can clarify your way. First, as you say, for all $\epsilon>0$ there exists $N>0$ such that $$||l_n-l_m||_\infty<\epsilon \qquad (n\geq N).$$ From this you get that for every $x\in[0,1]$, $$|l_n(x)-l_m(x)|<||l_n-l_m||_\infty<\epsilon\qquad (n\geq N).$$ Thus the sequence of real numbers $\{l_n(x)\}_{n\ge 1}$ is a Cauchy sequence on $\mathbb{R}$ for any $x\in [0,1]$. Hence the function $l:[0,1]\to \mathbb{R}$ given by $$l(x) = \lim_{n\to\infty} l_n(x) $$ is well-defined. Now your work is convince yourself that:

1. The function $l$ is a polynomial of degree less than or equal to 2, this is $l\in \textbf{P}$, and
2. There exist an $N'>0$ such that $||l_n-l||_\infty<\epsilon$ for all $n\geq N'$. Hint: Why $N'=N$ works? (the first $N$ you found).

Answer 3

All norms on a finite dimensional space are equivalent. (This is essentially the content of @geetha290krm's comment.)

In particular, the norm $\| \cdot \|_\infty$ in the question and the norm $\|x \mapsto a x^2+bx+c\|_c = \|(a,b,c)\|_\infty$ are all equivalent.

Hence if $l_n$ is Cauchy in $\| \cdot \|_\infty$, then the sequence is also Cauchy in $\| \cdot \|_c$, and so there are $a,b,c$ such that $(a_n,b_n,c_n) \to (a,b,c)$. Consequently, with $l(x) = a x^2+bx+c$, we see that $\|l_n -l \|_\infty \to 0$.