If \(f\) is \(C^2\) on an open interval, prove that \[ \lim_{h\to0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2} =f''(x). \]
Solution
Fix \(x\) in the open interval.
1. For every sufficiently small nonzero \(h\), there are points\(\xi_h^+\) between \(x\) and \(x+h\) and \(\xi_h^-\) between \(x\) and \(x-h\) suchthat\[ f(x+h) = f(x)+hf'(x)+\frac{h^2}{2}f''(\xi_h^+) \]and\[ f(x-h) = f(x)-hf'(x)+\frac{h^2}{2}f''(\xi_h^-). \]
Because the interval is open, \(x\pm h\) remain in it for all sufficiently small \(h\). Apply Taylor’s theorem with Lagrange remainder to \(f\) at \(x\), once at the point \(x+h\) and once at the point \(x-h\).
2. For such \(h\),\[ \frac{f(x+h)-2f(x)+f(x-h)}{h^2} = \frac{ f''(\xi_h^+)+f''(\xi_h^-) }{2}. \]
1. The constant terms leave\(2f(x)\) and the linear terms \(hf'(x)\) and \(-hf'(x)\) cancel. Subtract\(2f(x)\) and divide by \(h^2\).
3. As \(h\to0\),\[ \xi_h^+\longrightarrow x \qquad\text{and}\qquad \xi_h^-\longrightarrow x. \]
Each \(\xi_h^\pm\) lies between \(x\) and \(x\pm h\), so \[ \abs{\xi_h^\pm-x}\leq\abs h. \] The right side tends to zero.
4. One has\[ \lim_{h\to0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2} = f''(x). \]
Since \(f\) is \(C^2\), the function \(f''\) is continuous. Step 3 therefore gives \[ f''(\xi_h^+)\longrightarrow f''(x) \qquad\text{and}\qquad f''(\xi_h^-)\longrightarrow f''(x). \] Apply these limits to the identity in step 2.
5. Q.E.D.
Step 4 is the required limit.