$$f\in C^{2}(\mathbb{R})\textrm{ such that }\left | f''(x) \right |\leq M\: \: \forall x\in \mathbb{R}.\\ \textrm{Prove that }\frac{\left | f(x+h)+f(x-h)-2f(x) \right |}{h^{2}}\leq M\: \: \forall x\in \mathbb{R}$$ I have tried mean value theorem and other standard analysis results but have made no progress. Kindly help.
Asked
Active
Viewed 160 times
2 Answers
1
Hint: $\int_x^{x+h}f'(t)dt-\int_x^{x+h}f'(t-h)dt=f(x+h)+f(x-h)-2f(x)$. Apply MVT to $f'(t)-f'(t-h)$.
Kavi Rama Murthy
- 311,013
1
For fixed $x$ consider the function $$ g(h) = f(x+h) + f(x-h) - 2f(x) \, . $$ Then use Taylor's theorem in the form $$ g(h) = g(0) + hg'(0) + \frac 12 h^2 g''(c) $$ for some $c$ between $0$ and $h$, and note that $g(0) = g'(0) = 0$. This gives $$ \frac{f(x+h) + f(x-h) - 2f(x)}{h^2} = \frac {f''(x+c) + f''(x-c)}2 $$ and the desired estimate follows immediately.
Remarks:
- Only the existence of the second derivative of $f$ is needed for this conclusion, not its continuity.
- One even has $$ \frac{f(x+h) + f(x-h) - 2f(x)}{h^2} = f''(\xi) $$ for some $\xi \in (x-h, x+h)$. That follows from above calculations and the mean value property for derivatives, or from the general result Is there a mean value theorem for higher order differences?.
Martin R
- 113,040