Tutorial 2 Workings

5. Consider the space of continuous differentiable functions $C^1([a,b])$ with the $C^1$-norm

$$\Vert f \Vert=\sup_{x \in [a,b]}|f(x)|+\sup_{x \in [a,b]}|f'(x)|$$

Prove that $C^1([a,b])$ is a Banach space with respect to the given norm.

Let $(f_n)$ be a Cauchy sequence in the space $C^1([a,b])$, then for any $\varepsilon>0$, there exists a positive integer $N$ such that whenever $n,m \geqslant N$, we have that $\Vert f_n-f_m \Vert<\varepsilon$.

Note that if we take any $\varepsilon>0$ to obtain the corresponding $N$, and fix any $x \in X$, we have that

$$|f_n(x)-f_m(x)| \leqslant \sup_{x \in [a,b]}|f_n(x)-f_m(x)| \leqslant \Vert f_n-f_m \Vert < \varepsilon.$$

Since the inequality above holds for all $x \in [a,b]$, it follows that $f_n$ converges uniformly to a function $f$.

Using the same argument (replacing $f_n$ and $f_m$ by $f_n'$ and $f_m'$ respectively), we also see that the sequence $(f_n')$ obtained by taking the first derivative of every term of the sequence $(f_n)$ converges uniformly to a function $g$.

Due to the uniform convergence of $(f_n)$ and $(f_n')$, the limit $f$ is continuous and differentiable with $f'=g$.

We claim that $(f_n)$ converges to $f$. Let $\varepsilon>0$ be arbitrary. Applying the uniform convergence of $(f_n)$ and $(f_n')$, choose $N_1,N_2$ such that $\sup_{x \in [a,b]}|f(x)-f_n(x)|<\varepsilon/2$ whenever $n \geqslant N_1$ and $\sup_{x \in [a,b]}|f'(x)-f_n'(x)|<\varepsilon/2$ whenever $n \geqslant N_2$. It follows that when $n \geqslant \max\{N_1,N_2\}$, we have that $\Vert f-f_n \Vert = \sup_{x \in [a,b]}|f'(x)-f_n'(x)|+\sup_{x \in [a,b]}|f'(x)-f_n'(x)|<\varepsilon/2+\varepsilon/2=\varepsilon$. Therefore, $C^1([a,b])$ is a Banach space under the given norm.

6. Let $Y$ be a closed subspace of a normed linear space $(X, \Vert \cdot \Vert)$. Let $X/Y$ denote the quotient space (elements of $X/Y$ are additive cosets.) For $x + Y \in X/Y$ define the quotient norm $\Vert \cdot \Vert_*$ by

$$\Vert x + Y \Vert_* = \inf_{y \in Y}\Vert x-y \Vert$$

Show that $\Vert \cdot \Vert_*$ is a norm on $X/Y$. Also, if $X$ is a Banach space, show that $X/Y$ is a Banach space under the quotient norm.

Since $\Vert \cdot \Vert$ is positive, the infimum of all possible values of $\Vert x-y \Vert$ amongst all $y \in Y$ is also not smaller than zero, so $\Vert \cdot \Vert_*$ is positive.

Note that $\Vert x+Y \Vert_* =\inf_{y \in Y}\Vert x-y \Vert=0$ if and only if $\Vert x-y' \Vert=0$ for some $y' \in Y$ due to the positivity of $\Vert \cdot \Vert$. This happens if and only if $x = y'$ due to the definiteness of $\Vert \cdot \Vert$, which means that $x \in Y$, so $x + Y$ is the zero vector in the quotient space, thereby showing that $\Vert \cdot \Vert_*$ is definite.

Let $s$ be a scalar. We see that $\Vert sx + Y \Vert_* = \inf_{y \in Y}\Vert sx-y \Vert=\inf_{y \in Y}\Vert sx-sy \Vert = |s|\inf_{y \in Y}\Vert x-y \Vert=|s|\Vert x+Y \Vert_*$. Note that the second equality applies the fact that a vector subspace is closed under scalar multiplication, whereas the third equality applies the absolute homogeneity of $\Vert \cdot \Vert$. This proves the absolute homogeneity of $\Vert \cdot \Vert_*$.

As for the triangle inequality, we see that for $x_1,x_2 \in X$, we have that $\Vert x_1+x_2+Y \Vert_*=\inf_{y \in Y} \Vert x_1+x_2-y \Vert=\inf_{y \in Y} \Vert x_1-y+x_2-y \Vert \leqslant \inf_{y \in Y}\Vert x_1-y \Vert + \inf_{y \in Y} \Vert x_2-y \Vert = \Vert x_1+Y \Vert_* + \Vert x_2+Y \Vert_*$. Note that the second equality applies the fact that a vector subspace is closed under vector addition. This proves the triangle inequality for $\Vert \cdot \Vert_*$ .

This concludes the proof that $\Vert \cdot \Vert_*$ is a norm.

For an argument that shows that $X/Y$ is a Banach space under the quotient norm, see here.