Application of the Arzela-Ascoli theorem

$\textbf{Application of the Arzel{\'a}-Ascoli Theorem:}$

Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on [0,1] satisfying $\int_0^1 f_n^2(t)dt \leq M$ $\forall n\in{\mathbb{N}}$ and some $M \geq 0$. Define $g_n(x) = \int_0^xf_n(t)dt$. Show that $\exists \{g_{n_k}\} \subseteq \{g_n\}$ s.t. $g_{n_k} \longrightarrow g$ uniformly.

$\textit{Recall}$ : The Arzel{\'a}-Ascoli theorem states that if $\{g_n\}$ is an equicontinuous sequence of real-valued functions defined on a compact metric space that are uniformly bounded: $|g_n(x)| \leq M$ $\forall n$ and some $M \geq 0$, then, $\exists \{g_{n_k}\}$ s.t. $g_{n_k} \longrightarrow g\in{C(X)}$ uniformly.

$([0,1],|\cdot | )$ is a compact metric space...check. So, if we can show that $\{g_n\}$ is uniformly bounded and equicontinuous, then the Arzel{\'a}-Ascoli theorem will apply and give us the desired conclusion.
We now want to show that $\{g_n\}$ is uniformly bounded: By assumption, $f_n\in{L^2[0,1]}$. Let $h=1$ on $[0,1]$. Then H{\"o}lder's inequality with p=q=2 (Cauchy-Schwarz) gives us that $$||f_n \cdot 1||_1 \leq ||f_n||_2 \cdot ||1||_2$$ $$\int_0^1|f_n \cdot 1| \leq \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}} \left(\int_0^1|1^2|\right)^{\frac{1}{2}} = \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}}$$ Squaring both sides, $$\left(\int_0^1|f_n|\right)^2 \leq \int_0^1|f_n|^2 \leq M$$ Thus, $f\in{L^1[0,1]}$. Note that we actually showed that $L^p[a,b] \subseteq L^1[a,b] \quad \forall 1 \leq p < \infty$. To compare these inequalities with the $g_n$ terms, $$|g_n|^2=\left|\int_0^xf_n\right|^2 \leq \left(\int_0^x|f_n|\right)^2 \leq \left(\int_0^1|f_n|\right)^2 \leq M$$ $$|g_n|^2 \leq M \Longrightarrow |g_n| \leq \sqrt{M}$$ Thus, $\{g_n\}$ is a uniformly bounded sequence...check.
Now consider $0\leq x < z \leq 1$. $$|g_n(x)-g_n(z)|=\left|\int_0^xf_n - \int_0^z f_n \right|=\left|\int_x^zf_n\right|\leq \int_x^z|f_n \cdot 1|$$ Using H{\"o}lder's inequality again, $$\leq \left(\int_x^z|f_n|^2\right)^{\frac{1}{2}}\left(\int_x^z|1|^2\right)^{\frac{1}{2}} = \left(\int_x^z|f_n|^2\right)^{\frac{1}{2}}\sqrt{|x-z|}\leq \left(\int_0^1|f_n|^2\right)^{\frac{1}{2}}\sqrt{|x-z|} $$ $$ \leq \sqrt[4]{M}\sqrt{|x-z|}$$ So, $|g_n(x)-g_n(z)| \leq \sqrt[4]{M}\sqrt{|x-z|}$ which gives us a Lipschitz condition on $g_n$ $\forall n$. In the $\epsilon-\delta$ criterion for equicontinuity, if we choose $\delta \leq \frac{\epsilon^2}{\sqrt{M}}$ then $|g_n(x)-g_n(z)| \leq \epsilon$ making $\{g_n\}$ uniformly equicontinuous.