1. ## tayler's theorem

Let f : V W be C2, where V and W are normed vector spaces. Define

M_0 = sup{f(x)∥_w : x V}

M_1 = sup{f’(x)∥_L(v,w) : xV}

M_2 = sup{f”(x)∥_L2(v,w) : xV}

Assume that these are all finite. Show that

M_1^2 ≤ 4M_0M_2.

[Hint: use Taylor’s theorem to show that for any xV and vV, and any s>0,

f’(x)(v)W ≤ (f(x+sv) – f(x)W)/s + (s/2)M_2,

And so deduce f’(x)L(v,w) ≤ 2M_0/s + sM_2/2. Choose s carefully.]

2. It will help if you understand the scalar case before attempting the vector problem. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is a $C^2$-function. Let $M_0 = \sup|f(x)|$, $M_1 = \sup|f'(x)|$, $M_2 = \sup|f''(x)|$, supposing that these are all finite. Taylor's theorem says that for any $x\in\mathbb{R}$ and s>0, $f(x+s) = f(x) + sf'(x) + \tfrac12s^2f''(\xi)$ for some $\xi$ between x and x+s. Therefore $f'(x) = \tfrac1s\bigl(f(x+s)-f(x)\bigr) - \tfrac s2f''(\xi)$. By the triangle inequality, $|f'(x)| \leqslant \tfrac1s|f(x+s)-f(x)| + \tfrac12s|f''(\xi)| \leqslant \tfrac2sM_0 + \tfrac s2M_2$.
Now let $s = 2\sqrt{M_0/M_2}$. The previous inequality becomes $|f'(x)|\leqslant 2\sqrt{M_0M_2}$. Square both sides and take the sup over x to get $M_1^2\leqslant 4M_0M_2$.