Complete subspace

[Papapetros Vaggelis]

Consider the \(\displaystyle{\mathbb{R}}\) - space \(\displaystyle{\left(C(\left[0,1\right]),+,\cdot\right)}\) equipped

with the norm \(\displaystyle{||\cdot||_{\infty}:C(\left[0,1\right])\longrightarrow \mathbb{R}\,,f\mapsto ||f||_{\infty}=\max\,\left\{\left|f(x)\right|: 0\leq x\leq 1\right\}}\) .

Prove that the \(\displaystyle{\mathbb{R}}\) -subspace \(\displaystyle{Y=\left\{f\in C(\left[0,1\right]): f(0)=0\right\}}\) of

\(\displaystyle{\left(C(\left[0,1\right]),||\cdot||_{\infty}\right)}\) is complete.

[Tolaso J Kos]

Consider the \(\displaystyle{\mathbb{R}}\) - space \(\displaystyle{\left(C(\left[0,1\right]),+,\cdot\right)}\) equipped

with the norm \(\displaystyle{||\cdot||_{\infty}:C(\left[0,1\right])\longrightarrow \mathbb{R}\,,f\mapsto ||f||_{\infty}=\max\,\left\{\left|f(x)\right|: 0\leq x\leq 1\right\}}\) .

Prove that the \(\displaystyle{\mathbb{R}}\) -subspace \(\displaystyle{Y=\left\{f\in C(\left[0,1\right]): f(0)=0\right\}}\) of

\(\displaystyle{\left(C(\left[0,1\right]),||\cdot||_{\infty}\right)}\) is complete.

It is sufficient to prove that the subspace \( Y \) is closed. Let \( f \in \bar{Y} \). Then there exists a sequence , say \( f_n , \;\; n \in \mathbb{N} \) that \( f_n \xrightarrow{||\cdot||_{\infty}} f \). That is:

$$(\forall \epsilon>0 ) (\exists n_0 \in \mathbb{N}) (n\geq n_0) \Rightarrow ||f_n-f||_{\infty}<\epsilon$$

Hence:

$$(\forall \epsilon>0 ) (\exists n_0 \in \mathbb{N}) (n\geq n_0) (\forall x \in [0, 1]): |f_n(x)-f(x)|\leq ||f_n-f||_{\infty} < \epsilon$$

Therefore \( f_n \) converges uniformly to \( f\) , hence \( f \) continuous and \( f_n(0) \rightarrow f(0) \Rightarrow f(0)=0 \). That is \( f \in Y \) which is what we want.

[Papapetros Vaggelis]

Hint

The \(\displaystyle{\mathbb{R}}\) normed space \(\displaystyle{\left(C(\left[0,1\right]),||\cdot||_{\infty}\right)}\)

is a \(\displaystyle{\rm{Banach}}\) space.