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Let $\mathcal{L}$ be the space of real-valued Lipschitz functions of order 1 defined on $[0,1]$. That is, the class of functions $f$ such that $$ \sup\limits_{(x,y)\in[0,1]\times[0,1], x\neq y} \dfrac{\vert f(x)-f(y)\vert}{\vert x-y\vert} =K(f)<\infty. $$ Given $\Vert f\Vert_1=\sup\limits_{0\leq t\leq 1}\vert f(t)\vert +K(f)=\Vert f\Vert + K(f)$ is a norm on $\mathcal{L}$.

Show that $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is a Banach spaces.

To show $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is a Banach spaces, I use the definition of Banach spaces, i.e. $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is complete.

To show $\left(\mathcal{L},\Vert\cdot\Vert_1\right)$ is complete, I will prove every Cauchy sequence on $\mathcal{L}$ is converge, i.e.

Let $\{f_i\}_{i\in \mathbb{N}}$ be the sequence in $\mathcal{L}$. For all $\varepsilon>0$ there exist $N\in \mathbb{N}$ such that for all natural number $m,n>N$ $$\Vert f_m-f_n\Vert_1<\varepsilon.$$

Proof.

Let $\varepsilon>0$.

\begin{align*} \Vert f_m-f_n\Vert_1 &=\sup\limits_{0\leq t\leq 1}\vert f_m(t)-f_n(t)\vert +K(f_m-f_n)\\ &\leq \sup\limits_{0\leq t\leq 1}\vert f_m(t)\vert + \sup\limits_{0\leq t\leq 1}\vert f_n(t)\vert +K(f_m-f_n)\\ \end{align*}

Now, I don't know (don't have an idea) how to conclude less than $\varepsilon$. Any hint to prove it?

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1 Answer 1

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Let $\epsilon >0$. There exists $N$ such that $\sup _t |f_n(t)-f_m(t)| +K(f_n-f_n)<\epsilon$ for all $n,m >N$. This implies that $\sup _t |f_n(t)-f_m(t)| <\epsilon$ for all $n,m >N$ and $K(f_n-f_m)<\epsilon$ for all $n,m >N$. $(f_n(t))$ is a Cauchy sequence so $f(t)=\lim f_n(t)$ exists for each $t$. Letting $m \to \infty$ in the inequality $|f_n(t)-f_m(t)| <\epsilon$ we get $|f_n(t)-f(t)| \leq \epsilon$ for $n >N$. Similarly, we can let $m \to \infty$ in $K(f_n-f_m) <\epsilon$ to get $k(f_n-f) \leq \epsilon$ for $ n>N$. Rest of the argument should now be clear.

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