Take the 2-minute tour ×

Let $D\subseteq \mathbb R^2$ be an open set and $P:D\rightarrow \mathbb R$ continuous. For $y$ fixed how to evaluate,

$$\displaystyle\lim_{h\to 0}\frac{1}{h}\int_{x}^{x+h}P(t, y)\ dt?$$

I know the answer, the limit above must be $P(x, y)$ but I can't see why.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Note first that $$ \frac{1}{h}\int_x^{x+h}P(t,h)dt-P(x,y)=\frac{1}{h}\int_x^{x+h}(P(t,y)-P(x,y))dt $$ for all $|h|$ small enough such that $[x,x+h]\times\{y\}$ or $[x+h,x]\times\{y\}$ is contained in $D$.

Fix $\epsilon>0$.

Now by continuity of $t\longmapsto P(t,y)$ at $x$, there exists $\delta>0$ such that $|P(t,y)-P(x,y)|\leq \epsilon$ for $|h|\leq \delta$.

Then $$ |\frac{1}{h}\int_x^{x+h}P(t,y)dt-P(x,y)|\leq \frac{1}{|h|}|\int_x^{x+h}|P(t,y)-P(x,y)|dt|\leq \frac{1}{|h|}|h|\epsilon=\epsilon $$ for all non zero $|h|\leq \delta$.

So $$ \lim_{h\rightarrow 0}\frac{1}{h}\int_x^{x+h}P(t,y)dt=P(x,y). $$

share|improve this answer

Your Answer


