# If f is thrice continuously differentiatable ( f is in $C^{3}$) and its third derivative is bounded, then show that

|$$\frac{f(x+h)+f(x-h)-2f(x)}{h^{2}}-f''(x)$$| is less than or equal to $$h^{3}$$ times some constant times the supremum of $$|f''(y)|$$ over y in R

$$f'''(x)$$ = $$lim_{h\rightarrow 0} \frac{f(x+3h)-f(x+2h)}{h^{3}}-\frac{2f(x+2h)-f(x+h)}{h^{3}}+\frac{f(x+h)-f(x)}{h^{3}}$$

$$f'''(x)$$= $$lim_{h\rightarrow 0} \frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^{3}}$$

not sure how to proceed

Hint: Taylor formulas yield that $$f(x \pm h)=f(x) \pm hf’(x)+\frac{h^2}{2}\int_0^1{(1-u)f’’(x \pm hu)du}$$.
• That is why I wrote it was a hint (you even asked how to proceed, not the full solution). You can figure out the solution from this, especially if you realize, say, that $f’’$ is Lipschitz continuous with constant $\sup_y |f’’’|$. Dec 18, 2018 at 23:47