Constructing a proof regarding continuity and differentiability of a 2-variables function

Question

The claim I am trying to prove is that:

Let a two-variable function $f(x,y)$ be continuous and differentiable at any point in $\mathbb{R}^2$. Then for any a ∈ $\mathbb{R}$,

$\lim_{h\to 0} \frac 1h \int_a^{a+h} f(h,x) dx = f(0,a)$.

Specifically, I would like to know how to tackle the part that ${h\to 0}$ incurs the differentiation and causes the first variable of $f$ to approach zero at the same time.

Should I deal with one variable after another (i.e., $g(x) := f(h,x)$), or use some other strategy?

Answer 1

We have $$ (1) \qquad \frac 1h \int_a^{a+h} f(h,x) dx - f(0,a) = \frac 1h \int_a^{a+h} (f(h,x) - f(0,a) )dx . $$ Using continuoity of $f(x,y) $ at $(0,a)$ we have that for any $\varepsilon>0 $ small there exists $h_0>0$ such that once $0<h<h_0$ then $$ | f(h,x) - f(0,a) | \leq \varepsilon \ \text{ for all } \ 0\leq h \leq h_0 , a\leq x\leq a+h. $$ Hence for $h<h_0$, from (1) we get $$ \left| \qquad \frac 1h \int_a^{a+h} f(h,x) dx - f(0,a) \right| \leq \frac{1}{h} \int_a^{a+h} |f(h,x) - f(0,a) | dx \leq \varepsilon . $$ Hence the result follows.

EDIT: my initial answer used Lagrang's intermediate value theorem, but what we need here is the continuity only.

Answer 2

Use the mean value theorem for integrals: If $g\colon [c,d] \to \mathbb{R}$ is continuous, then \begin{equation*} \int_c^d g(x) \; dx = (d-c)g(\zeta) \end{equation*} for some $\zeta \in [c,d]$. In your problem, consider $g(x)=f(h,x)$ and $[c,d]=[a,a+h]$ for a given $h$. I don't think that the differentiability hypothesis about $f$ is important, though.