When I came across the Morse Lemma, which is on page 5 of 'Morse Theory' by Milnor, I found one little calculus thing in its proof, in lemma 2.1 regarding the function $g_i$, which I fail to understand. It itself has nothing to do with Morse functions as far as I can see. It comes down to this :
If I have f which is C² ($R^n \rightarrow R$ and Morse), $h(t) = f(tx)$ and $t \in [0,1]$, then
$f(x) = h(1)-h(0) = \int_0^1 h'(t)dt = x \int_0^1 f'(tx)dt = \sum_{j}x_j \int_0^1 {\partial f \over \partial x_j} (tx) dt $.
Now $\int_0^1 {\partial f \over \partial x_j}(tx)dt$ must be C¹ , but I fail to see why exactly:
You would show this using the theorem on differentiation under an integral sign:
$t \rightarrow {\partial f \over \partial x_j}(tx)$ can be integrated by construction
$x \rightarrow{\partial f \over \partial x_j}(tx)$ is C¹ because f is C² is and tx is linear
$t \rightarrow {\partial^2 f\over \partial x_j\partial x_k}(tx)$ is continuous because f is C²
what I am left with, is whether ${\partial^2 f\over \partial x_j\partial x_k}(tx) $ is dominated by an integrable function for all x and t. For every x, ${\partial^2f \over \partial x_j\partial x_k} (tx)$ is bounded because f'' is continuous and t is from a compact set. But how do I know that $|{\partial^2f\over \partial x_j\partial x_k}(tx)|$ is dominated for all t AND x ??
Obviously there is an embarassingly little thing that I am not seeing here. That derivation under the integral sign is used, I have seen confirmed in a handout I found on stanford.edu, without any further comment. So it must be something extremely basic.
Thanks for any tips
The property you ask about (differentiability with respect to $x$) is local in $x$, so you can assume $x$ is in a bounded interval, and continuous functions on the real line are bounded on each bounded interval. See the bottom row of Table 2 here.