By what I can see from the picture, the expected call of the environments you define is of the form
\begin{theorem}{1}
Let $\mathscr{H}$ be a Hilbert space ...
\end{theorem}
and your output seems instead the result of
\begin{theorem}[Theorem 1]
Let $\mathscr{H}$ be a Hilbert space ...
\end{theorem}
which is wrong. Use the former syntax.
Note
I have doubts about your way of using LaTeX. Given those definitions, just one new environment suffices. You can call
\begin{theorem}{1}
Let $\mathscr{H}$ be a Hilbert space ...
\end{theorem}
\begin{theorem}[Corollary]{1}
In the situation of ...
\end{theorem}
and the output would be exactly the same, without the need for six almost identical definitions.
However, this is against the spirit of LaTeX, where manual numbering can and should be avoided. For theorem-like statements, LaTeX provides \newtheorem, which is improved by loading the amsthm package.
\documentclass{article}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{mathrsfs}
\theoremstyle{definition} % body font upright
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\begin{document}
\begin{theorem}\label{thm:projection}
Let $\mathscr{H}$ be a Hilbert space and $\mathscr{M}$ a closed
subspace of~$\mathscr{H}$. Moreover, let $x\in\mathscr{H}$. Then
there is a unique element $\hat{x}\in\mathscr{M}$ such that
\begin{equation}
\label{eq:projection}
\lVert x-\hat{x}\rVert = \inf_{y\in\mathscr{M}}\lVert x-y\rVert.
\end{equation}
Additionally, $\hat{x}\in\mathscr{M}$ and \eqref{eq:projection}~iff
$\hat{x}\in\mathscr{M}$ and $(x-\hat{x}\in\mathscr{M}^{\perp}$.
\end{theorem}
\begin{proof}
See \cite[p.~51]{BD91}.
\end{proof}
\begin{corollary}
In the situation of Theorem~\ref{thm:projection}, let $\mathcal{I}$
denote the identity map on $\mathscr{H}$. Then there is a unique map
$P_{\mathscr{M}}$ of $\mathscr{H}$ onto $\mathscr{M}$ such that
$\mathcal{I}-P_{\mathscr{M}}$ maps $\mathscr{H}$ onto $\mathscr{M}^{\perp}$.
\end{corollary}
\begin{thebibliography}{BD91}
\bibitem[BD91]{BD91} Whatever
\end{thebibliography}
\end{document}

\newtheorem? – egreg Oct 23 '15 at 20:04\begin{theorem}Content of theorem.\end{theorem}. In that case,Cis treated as the mandatory argument andontent of theorem.is treated as the body. The way you've defined the environment, it takes one optional and one required argument. So the syntax is\begin{theorem}[<optional>]{<required>}<contents>\end{theorem}. But, as @egreg says, this is not a good way to do it in any case. – cfr Oct 23 '15 at 20:19\begin{theorem}[Theorem 1] Let ...\end{theorem}, but it should be\begin{theorem}{1} Let ...\end{theorem}– egreg Oct 23 '15 at 20:20\(provide/re)newcommand(*),\(provide/re)newenvironment(*)etc. It may be helpful. – cfr Oct 23 '15 at 20:21