Prove that $\Gamma=\{\gamma:K\longrightarrow (Y, \| . \|) : \gamma\; \text{is continuous and} \; \gamma|_{K_0}=\gamma_0\}$ is a complete metric space.

Question

QUESTION: Let $(X, \|.\|)_X$ be a Banach space and $Y\subset X$ a subspace, which is itself a Banach space endowed with a norm $\|.\|_Y$ such that $\|y\|_X\leq \|x\|_Y$ for every $y\in Y$. Let $K$ be a compact metric space, $K_0\subset K$ a closed subset and $\gamma_0: K_0\longrightarrow (Y, \|.\|_Y)$ a continuous map. Prove that $$\Gamma=\left\{\gamma:K\longrightarrow (Y, \| . \|) : \gamma\; \text{is continuous and} \; \gamma|_{K_0}=\gamma_0\right\}$$ is a complete metric space endowed with the uniform distance $$d_{\Gamma}(\gamma_1, \gamma_2)=\displaystyle\max_{t\in K} \|\gamma_1(t)-\gamma_2(t)\|_Y,$$ with $\gamma_1, \gamma_2\in \Gamma$.

About show that is a metric space, I've done. I'm not sure if I showed properly that every Cauchy sequence converges in this space. Would someone check for me, please?

MY ATTEMPT:

Let $(\gamma_n)$ be a Cauchy sequence in $\Gamma$. Then, for every $\epsilon >0$ given there exists $M=M(\epsilon)\in \mathbb{N}$ such that $$d_{\Gamma}(\gamma_n, \gamma_m)=\displaystyle\max_{t\in K} \| \gamma_n(t) - \gamma_m(t)\|_Y <\epsilon$$ for any $m,n\geq M$, which implies that $$\| \gamma_n(t), \gamma_m(t)\|_Y <\epsilon, \; \forall t\in K\; \text{and}\; m,n\geq M.$$ Thereby, $(\gamma_n(t))$ is a Cauchy sequence in $(Y, \|.\|_Y)$, however $Y$ is a Banach space, thus there exist $y\in Y$ such that $\gamma_n(t)\rightarrow y\in Y\tag{I}$

Defining, $$\begin{array}{cccc} \overline{\gamma} \ : & \! K & \! \longrightarrow & \! (Y, \|.\|)_Y\\ & \! x & \! \longmapsto & \! \overline\gamma(x) \end{array}$$

Claim: $\gamma_n\rightarrow \overline{\gamma}\; \in \Gamma.$ Proof: Indeed, first of all note that $\gamma_n$ and $\overline{\gamma}$ are uniformly continuous in $K$. This is, one has respectively that

1. For all $\epsilon>0$, there exists $\delta(n)>0$ such that, for any $x,y\in K$, $d(x,y)<\delta(n) \implies d_{\Gamma}(\gamma_n(x), \gamma_n(y))< \frac{\epsilon}{4}.$
2. For all $\epsilon>0$, there exists $\delta>0$ such that, for any $x,y \in K$, $d(x,y)<\delta\implies d_{\Gamma}(\overline{\gamma}(x), \overline{\gamma}(y))< \frac{\epsilon}{4}.$

Now, choosing $N\in \mathbb{N}$ such that $N>n$ in $(1)$ and replacing $d(x,y)<\delta(N)$ when $N>n $ in $(2)$, it follows that

\begin{align*} d_{\Gamma}(\gamma_n, \overline{\gamma})\\ &=\displaystyle\max_{x\in K}\|\gamma_n(x) - \overline{\gamma}(x) \|_Y \\ &=\displaystyle\max_{x\in K}\|\gamma_n(x) - \overline{\gamma}(x)+\overline{\gamma}(x)-\overline{\gamma}(x) + \overline{\gamma}(y)\\ &-\overline{\gamma}(y)+\gamma_n(x)-\gamma_n(x)+\gamma_n(y)-\gamma_n(y) \|_Y\\ &\leq \displaystyle\max_{x\in K}\|2 \gamma_n(x) - \gamma_n(y)\|_Y + \displaystyle\max_{x\in K}\| \gamma_n(x) -\gamma_n(y)\|_Y\\ & + \displaystyle\max_{x\in K}\|\overline{\gamma}(x) - \overline{\gamma}(y) \|_Y + \displaystyle\max_{x\in K}\|-2\overline{\gamma}(x) + \overline{\gamma}(y) \|_Y\\ &\leq\|\gamma_n(x)\|_Y \displaystyle\max_{x\in K}\|\gamma_n(x) - \gamma_n(y)\|_Y + \displaystyle\max_{x\in K}\| \gamma_n(x) - \gamma_n(y) \|_Y \\ &+ \displaystyle\max_{x\in K}\| \overline{\gamma}(x)-\overline{\gamma}(y)\|_Y + \|\overline{\gamma}(x)\|_Y \displaystyle\max_{x\in K}\|\overline{\gamma}(x) - \overline{\gamma}(y)\|_Y \\ &< \|\gamma_n(x)\|_Y \frac{\epsilon}{4}+ \frac{\epsilon}{4} + \frac{\epsilon}{4} + \|\overline{\gamma}(x)\|_Y \frac{\epsilon}{4}\\ &\overset{\mathrm{(I)}}{=} \|y\|\frac{\epsilon}{4} + \frac{\epsilon}{2} + \|\overline{\gamma}(x)\| \frac{\epsilon}{4}\\ \end{align*}

One knows that $\|\overline{\gamma}(x)\|$ is limited in $Y$ once $\overline{\gamma}$ is a continuous function defined in a compact set. Therefore, one takes the limit as $\epsilon \rightarrow 0$ to conclude that $d_{\Gamma}(\gamma_n, \overline{\gamma})=0$.

Answer 1

To see $\Gamma$ is a complete, note $\Gamma$ is contained in $C(K,Y)$, which is a complete metric space as $K$ is compact and $Y$ is complete. Therefore it suffices to show $\Gamma$ is a closed subset of $C(K,Y)$.

Pick $(\gamma_n)$ any convergent sequence in $\Gamma$, $\gamma_n \to \gamma \in C(K,Y)$, it is clear that $\gamma(x) = \gamma_0(x)$ for every $x \in K_0$. So $\gamma \in \Gamma$ and $\Gamma$ is closed in $C(K,Y)$.