Jtdylktuy

Let $Lambda_{alpha}([0,1])$ be the space of $alpha$ Holder continuous functions on $[0,1]$ with the norm:
$$|f|_{Lambda_{alpha}} = |f(0)| + sup_{x,y in [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|^{alpha}}$$
and consider the subspace $lambda_{alpha}$ given by
$$frac{|f(x) - f(y)|}{|x-y|^{alpha}} rightarrow 0 text{ as } x rightarrow y quad forall , yin[0,1]$$

I'm having trouble showing that for $alpha < 1$ this is an infinite dimensional closed subspace of $Lambda_{alpha}([0,1])$. I showed it was a subspace (as a vector space). Is there some theorem I should be using here? I started with a cauchy sequence in $lambda_{alpha}$ but I'm having trouble saying anything about its limit -- other than it lives in $Lambda_{alpha}$.

EDIT: Does this work?

Let $(f_n)$ be a Cauchy sequence in $lambda_{alpha}$.
begin{align}
frac{|f(x)-f(y)|}{|x-y|^{alpha}} & = frac{|lim_{nrightarrow infty}f_{n}(x)-lim_{nrightarrow infty}f_{n}(y)|}{|x-y|^{alpha}} \
& = lim_{n rightarrow infty}frac{|f_{n}(x)-f_{n}(y)|}{|x-y|^{alpha}} rightarrow 0 text{ as } x rightarrow y
end{align}

In your edit, you can insert a zero:
begin{align}
frac{|f(x)-f(y)|}{|x-y|^{alpha}} &
le frac{|f(x)-f_{n}(x)|}{|x-y|^{alpha}}+frac{|f_n(x)-f_{n}(y)|}{|x-y|^{alpha}} +frac{|f_n(y)-f(x)|}{|x-y|^{alpha}}
end{align}
Choose $epsilon>0$. Then the first and last term are less than $epsilon/3$ for all $n$ large enough. Fix such an $n$. Then the second term is less than $epsilon/3$ for $|x-y|$ small enough. This shows that
$$ frac{|f(x)-f(y)|}{|x-y|^{alpha}}le
epsilon
$$
for all $y$ close to $x$. This is the claim.