It is not clear what you mean by "formatting to be on the left." If you mean equation numbers on the left, you can use leqno option, if you want the displayed equations not be centered but left-aligned (with some margin), you can use the fleqn option with the documentclass, as in the below code.
You use the align environment but have not set any alignment points. I assumed you want alignment at leq signs right after the line breaks; hence the example below.
If your proof ends with an equation, use qedhere at the end for better placement of the end-of-proof sign.
Your inline equation after (1) is too long to fit on a line. Consider rewritingI put it in an align* environment.
Equations are part of the sentences, so they need proper punctuation. I added a few missing ones.
\documentclass[fleqn,leqno]{article} \usepackage{amsmath} \usepackage{amsthm} \begin{document} \begin{proof} First we prove \eqref{E-alpha-1}. \begin{align} \left|\mathbf{E}_{\alpha,1}^{(\beta)}( t) v\right|_{\mu} = \sum_{j=1}^{\infty}\lambda_j^{\mu}\langle(E_{\alpha,1}^{(\beta)}(t)v,e_j\rangle^2. \end{align} Note that $\langle \begin{align*} \langle(\mathbf{E}_{\alpha,1}^{(\beta)}t)v,e_j\rangle =&= \sum_{k=0}^{\infty}\frac{\langle (-A)^{\beta k}t^{\alpha k}v,e_j\rangle}{\Gamma(\alpha k+1)} =\\&= \sum_{k=0}^{\infty}\frac{\langle (-\lambda_j)^{\beta k}t^{\alpha k}v,e_j\rangle}{\Gamma(\alpha k+1)} = E_{\alpha,1}(-\lambda_j^{\beta} t^{\alpha})\langle v, e_j \rangle$.\rangle Using\end{align*} using that $A^{\beta k}e_j = \lambda_j^{\beta k} e_j$ whenever $e_j$ are the eigenvectors of $A$. Next we use the result that $|E_{\alpha,1}(z)|^2\leq \frac{C}{(1+|z|)^2}$. Thus, \begin{align} |\mathbf{E}_{\alpha,1}^{(\beta)}( t) v|_{\mu}^2 &= \sum_{j=1}^{\infty}\lambda_j^{\mu}\langle(E_{\alpha,1}^{(\beta)}(t)v,e_j\rangle^2 = \sum_{j=1}^{\infty}\lambda_j^{\mu}\langle v, e_j \rangle^2 |E_{\alpha,1}(-\lambda_j^{\beta} t^{\alpha})|^2 \\ & \leq \sum_{j=1}^{\infty}\lambda_j^{\mu}\langle v, e_j \rangle^2 \frac{1}{(1+\lambda_j^{\beta}t^{\alpha})^2}\\ & \leq C t^{-\frac{\alpha}{\beta}(\mu - \nu)} \sum_{j=1}^{\infty}\frac{(\lambda_j^\beta t^{\alpha})^{\frac{\mu-\nu}{\beta}}}{(1+\lambda_j^{\beta}t^{\alpha})^2}\lambda^{\nu}_j\langle v, e_j \rangle^2 \\ & \leq Ct^{-\frac{\alpha(\mu-\nu)}{\beta}}\sum_{j=1}^{\infty}\lambda_j^{\nu}\langle v, e_j\rangle^2 \leq Ct^{-\frac{\alpha(\mu-\nu)}{\beta}}|v|_{\nu}^2. \qedhere \end{align} \end{proof} \end{document}

