How can I make an align equation look neater

Perhaps you had something like the code below in mind. I removed the equation numbering, but if you want to bring it back, simply replace align* with align. The fleqn option allows you to left-align equations. Also, since my crystal ball seems to have gone missing, I included in the preamble only what your formulas required.

\documentclass[12pt,fleqn]{article} % fleqn = формулы прижаты к левому краю \usepackage{amsmath,amssymb,amsthm} \newtheorem{theorem}{Theorem} \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 \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)} \\[6pt] &= \sum_{k=0}^{\infty} \frac{\langle (-\lambda_j)^{\beta k} t^{\alpha k} v, e_j \rangle} {\Gamma(\alpha k+1)} \\[6pt] &= E_{\alpha,1}(-\lambda_j^{\beta} t^{\alpha}) \langle v, e_j \rangle . \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 \[ |E_{\alpha,1}(z)|^2 \le \frac{C}{(1+|z|)^2}. \] 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 \\[6pt] &= \sum_{j=1}^{\infty} \lambda_j^{\mu} \langle v, e_j \rangle^2 |E_{\alpha,1}(-\lambda_j^{\beta} t^{\alpha})|^2 \\[6pt] &\le \sum_{j=1}^{\infty} \lambda_j^{\mu} \langle v, e_j \rangle^2 \frac{1}{(1+\lambda_j^{\beta} t^{\alpha})^2} \\[6pt] &\le 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 \\[6pt] &\le C t^{-\frac{\alpha(\mu-\nu)}{\beta}} \sum_{j=1}^{\infty} \lambda_j^{\nu} \langle v, e_j \rangle^2 \\[6pt] &\le C t^{-\frac{\alpha(\mu-\nu)}{\beta}} |v|_{\nu}^2 . \end{align*} \end{proof} \end{document} 

Some pointers:

1. 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.

2. 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.

3. If your proof ends with an equation, use qedhere at the end for better placement of the end-of-proof sign.

4. Your inline equation after (1) is too long to fit on a line. I put it in an align* environment.

5. Equations are part of 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 \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 \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}