How does curvature change under perturbations of a Riemannian metric? Let $M$ be a compact subset of $\mathbb R^2$ with smooth boundary, and let $g$ be a Riemannian metric on $M$.  If $g'$ is another Riemannian metric which is "close" to $g$, then they should have almost identical curvature profiles.  I would like to prove a concrete estimate on the total difference of their curvatures in terms of the distance of $g'$ to $g$.  Before I state the question precisely, I need to introduce some notation.
Write $\operatorname{Sym}$ for the space of symmetric $2\times 2$ real matrices, and let $\operatorname{SPD} \subseteq \operatorname{Sym}$ be those matrices which are also positive-definite.  Consider the function space $\Omega = C^2(M, \operatorname{SPD})$.  Denote partial derivatives of $g_{ij} \in \Omega$ by additional subscripts following a comma, so that $\tfrac{\partial}{\partial x^k} g_{ij} = g_{ij,k}$, et cetera.  Endow the space $\Omega$ with the norm $$\|g\| = \sup_{x \in M} \max_{i,j,k,l} \left\{|g_{ij}(x)|, |g_{ij,k}(x)|, |g_{ij,kl}(x)| \right\},$$ so that it has the structure of an open cone within the Banach space $C^2(M, \operatorname{Sym})$.
Each $g \in \Omega$ defines a Riemannian structure on $M$ via the inner product $\langle v, g(x) v' \rangle$ for $v, v' \in T_x M$.  Let $K(g,x)$ be the scalar curvature of the metric $g$ at the point $x \in M$.
What I want to prove:  For each $g \in \Omega$, there exist constants $C$ and $\epsilon$ so that if $g' \in \Omega$ with $\|g - g'\| < \epsilon$, then $$\sup_{x \in M} \left| K(g,x) - K(g',x) \right| \le C\|g- g'\|.$$
My current approach to this is quite clunky, and involves calculating everything directly from the Christoffel symbols of the metrics.  Is there a better, more geometric approach to this than brute force calculations?
I'm sure this type of lemma is well known to geometric analysts.  Is a proof of a similar result written down somewhere?
 A: The coordinate-dependent approaches mentioned in other posts are the quickest way to proceed here, but if you want a more coordinate-independent approach, one can flow from one metric g to the next g' (e.g. by using the line segment $t \mapsto (1-t) g + t g'$) and using the standard formulae for the first variation of curvature, as can be found for instance in my blog post
http://terrytao.wordpress.com/2008/03/28/285g-lecture-1-ricci-flow/
A: It is a fact (due to Riemann I believe) that in normal coordinates, the Taylor expansion of $g_{ij}$ is $\delta_{ij}+\frac{1}{3}R_{ikjl}x^k x^l+O(||x||^3)$, where $R_{ikjl}$ are components of the $(4,0)$ curvature tensor.  In dimension $2$ the tensor reduces to scalar curvature. Thus
curvature is the second derivative of the metric in normal coordinates. In your setup you insist on global coordinates coming from the ambient Euclidean plane, so you need to take into account the coordinate change from the normal coordinates (defined locally) and global Euclidean coordinates. It seems to me that compactness of $M$ gives a bound on such coordinate change, a bound that depends on $g$. 
EDIT: After seeing comments by Will Jagy and Willie Wong, I realized that I misread the question and it makes no sense as stated. Still I will leave my answer in the hope that it would help to the questioner.
A: This is a straightforward consequence of the fact that $K(x)$ is a continuous function of $g(x)$, $\partial g(x)$, and $\partial^2(g)$.
A: Elaborating on Deane Yang's answer and Willie Wong's comment: Since
$M^{2}\subset\mathbb{R}^{2}$ is a $C^{\infty}$ submanifold with boundary, the
Euclidean coordinates are global. Generally, if $M^{n}$ is a compact manifold
with boundary, we can cover it by a finite number of charts $\{x^{i}\}$, where
for any $C^{\infty}$ metric $g$ the functions $g^{ij}$ and $\partial^{\alpha
}g_{ij}$ are bounded (depending on $g$ and $|\alpha|$) and where $\alpha$ is a
multi-index with $|\alpha|\geq0$.
The scalar curvature $R_{g}$ (twice the Gauss curvature $K$ if $n=2$) is
$$
R_{g}=g^{jk}(\partial_{\ell}\Gamma_{jk}^{\ell}-\partial_{j}\Gamma_{\ell
k}^{\ell}+\Gamma_{jk}^{p}\Gamma_{\ell p}^{\ell}-\Gamma_{\ell k}^{p}\Gamma
_{jp}^{\ell})=(g^{-1})^{2}\ast\partial^{2}g+(g^{-1})^{3}\ast(\partial g)^{2}
$$
since the Christoffel symbols have the form $\Gamma=g^{-1}\ast\partial g$,
where $\partial^{k}g$ denotes some $k$-th partial derivative of $g_{ij}$ and
where $\ast$ denotes a linear combination of products while summing over
repeated indices. From the formula for $R$ we have for metrics $g,g^{\prime}$,
\begin{align*}
& |R_{g}(x)-R_{g^{\prime}}(x)|\\
& \leq C(|g^{-1}|^{2}+|g^{\prime-1}|^{2})|\partial^{2}g-\partial^{2}g^{\prime
}|+C(|g^{-1}|^{4}+|g^{\prime-1}|^{4})(|\partial g|^{2}+|\partial g^{\prime
}|^{2})|g-g^{\prime}|\\
& +C(|g^{-1}|^{3}+|g^{\prime-1}|^{3})\{(|\partial^{2}g|+|\partial^{2}
g^{\prime}|)|g-g^{\prime}|+(|\partial g|+|\partial g^{\prime}|)\left\vert
\partial g-\partial g^{\prime}\right\vert \}
\end{align*}
since $|g^{-1}-g^{\prime-1}|\leq C(|g^{-1}|^{2}+|g^{\prime-1}|^{2}
)|g-g^{\prime}|$.
Let $\hat{\Omega}=C^{2}(M,\operatorname{Sym})$. Given $h\in\hat{\Omega}$,
define $||h||=\sup_{x\in M}\max_{i,j,k,\ell}\{|h_{ij}(x)|,|h_{ij,k}
(x)|,|h_{ij,k\ell}(x)|\}$. Then $|R_{g}(x)-R_{g^{\prime}}(x)|\leq
C||g-g^{\prime}||$, where $C$ depends on bounds on the inverses and the first
and second derivatives of $g$ and $g^{\prime}$.
Elaborating on Terence Tao's answer and Deane Yang's comment: One reason it is
convenient to compute in local coordinates $\{x^{i}\}$ is that $[\partial
_{i},\partial_{j}]=0$. So the expression for the Christoffel symbols has only
$3$ terms instead of the $6$ terms comprising the formula for $\nabla$:
$\Gamma_{ij}^{k}=\frac{1}{2}g^{k\ell}(\partial_{i}g_{j\ell}+\partial
_{j}g_{i\ell}-\partial_{\ell}g_{ij})$, which is symmetric in $i$ and $j$. With
$\frac{\partial}{\partial s}g_{ij}=v_{ij}$, the variation formula is easy to
compute: $\frac{\partial}{\partial s}\Gamma_{ij}^{k}=\frac{1}{2}g^{k\ell
}(\nabla_{i}v_{j\ell}+\nabla_{j}v_{i\ell}-\nabla_{\ell}v_{ij})$, since the
computation of this tensor formula at any point $p$ may be done in coordinates
where $\partial_{i}g_{jk}(p)=0$ (such as normal coordinates centered at $p$);
this enables us to convert $\partial_{i}$ to $\nabla_{i}$ and to ignore the
$\frac{\partial}{\partial s}g^{k\ell}$ term since it is multiplied by terms of
the form $\partial g$. Now the variation of the Riemann curvature tensor is
$\frac{\partial}{\partial s}R_{ijk}^{\ell}=\nabla_{i}(\dfrac{\partial
}{\partial s}\Gamma_{jk}^{\ell})-\nabla_{j}(\dfrac{\partial}{\partial s}
\Gamma_{ik}^{\ell})$ using the same trick of computing at the center $p$ of
normal coordinates and replacing $\partial$ by $\nabla$ (note that
$\frac{\partial}{\partial s}(\Gamma\ast\Gamma)=0$ at $p$ by the product rule);
the resulting formula is true in any coordinates since it is tensorial.
Generally, it is convenient to compute in local coordinates because it can be
done more or less mechanically. For example, if $\alpha$ is a $1$-form, then
$\nabla_{i}\nabla_{j}\alpha_{k}-\nabla_{j}\nabla_{i}\alpha_{k}=-R_{ijk}^{\ell
}\alpha_{\ell}$. One can remember this as the contraction of
$-\operatorname{Rm}$ and $\alpha$, where the lower indices $i,j,k$ on
$\operatorname{Rm}$ appear in the same order as the first term on the left.
Similarly, if $\beta$ is a $2$-tensor, then $\nabla_{i}\nabla_{j}\beta_{k\ell
}-\nabla_{j}\nabla_{i}\beta_{k\ell}=-R_{ijk}^{m}\beta_{m\ell}-R_{ij\ell}%
^{m}\beta_{km}$, where the the lower indices of $\operatorname{Rm}$ are $i,j$
and then either $k$ or $\ell$, with upper dummy index $m$ on
$\operatorname{Rm}$ also replacing either $k$ or $\ell$ on $\beta$.
