Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.