Let f be a function twice differentiable and with derivatives continuous on an interval $[a,b]$ containing $0$. Prove the following statement:

Assuming only $f''(0)$ exists, we have by Taylor (aka the MVT applied twice)

$$f(x)=f(0)+f'(0)x+ (f''(0)/2)x^2 + o(x^2).$$

Thus $(f(x)-f(0))/x = f'(0) +(f''(0)/2)x + o(x).$ Therefore

$$\frac{f'(x) - (f(x)-f(0))/x}{x} = \frac{f'(x) - f'(0) -(f''(0)/2)x + o(x))}{x}$$ $$ = \frac{f'(x) - f'(0)}{x} -f''(0)/2 + o(1).$$

As $x\to 0,$ this $\to f''(0)-f''(0)/2 = f''(0)/2,$ and we're done.

Tags:

Calculus

Real Analysis

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