Let $f:\left[a, b\right]\to\mathbb{R}$ be twice differentiable.

Suppose $f(a)=f(b)=f_{+}'(a)=f_{-}'(b)=0$ and $|f''(x)|\le M$.

Show that $|f(x)|\le \frac{M(b-a)^2}{16}$.

My try:

From taylor expansion we have $f(x)=\frac{f''(\xi _1)}{2}(x-a)^2$ and $f(x)=\frac{f''(\xi _2)}{2}(x-b)^2$.

Then $|f(x)|\le\frac{M}{2}(x-a)^2$ and $|f(x)|\le\frac{M}{2}(x-b)^2$

So $|f(x)|\le\min{(\frac{M}{2}(x-a)^2,\frac{M}{2}(x-b)^2)}\le\frac{M}{8}(b-a)^2$

But this conclusion is weaker than the aim.

And I thought through using $$f(c)=\frac{f''(\xi )}{2}(c-a)(c-b)$$ I can say$$|f(x)|\le\frac{M}{8}(b-a)^2$$ which only uses the condition that $f(a)=f(b)=0$ and is still weaker.

Any hints? Thank you in advance!


Let $c \in [a, b]$ be a point where $|f(x)|$ attains its maximum value on the interval. If $c=a$ or $c=b$ then $f$ is identically zero and we are done. Now assume $a < c < b$, then $f'(c) = 0$. We have to show that $|f(c)| \le \frac{M(b-a)^2}{16}$.

Case 1: If $a < c \le \frac{a+b}2$ then we can proceed as follows:

$f_{+}'(a) = f'(c) =0$ and $|f''(x)|\le M$ implies $$ |f'(x)| \le M \min(x-a, c-x ) \quad \text{for } a < x < c $$ and therefore, using $f(a) = 0$, $$ |f(c)| \le \int_a^c |f'(x)| \, dt \le M \int_a^c \min(x-a, c-x ) \, dt \\ = M \left( \int_a^{(a+c)/2} (x-a)\, dt + \int_{(a+c)/2}^c (c-x) \, dt \right) \\ = M \frac{(c-a)^2}8 \le M \frac{(b-a)^2}{16} \, . $$

Case 2: If $\frac{a+b}2 \le c < b$ then we can estimate $$ |f(c)| \le \int_c^b |f'(x)| \, dt $$ in the same way.


I thought the method I mentioned can be modified.

Let $f(x_0)=\displaystyle\max_{[a,b]} f$, we aim to prove $f(x_0)\le\frac{M}{16}(b-a)^2$.

Proof: According to the definition, we have $f'(x_0)=0$.

Without losing generality, we assume that $x_0\in[a,\frac{a+b}{2}]$

Because $|f'(x)|\le M$, we have $$f(x)\le \frac{1}{2}M(x-a)^2$$ In addition, $$f(x)\ge f(x_0)-\frac{1}{2}M(x-x_0)^2$$ which leads to $$f(x_0)\le\frac{M}{16}(b-a)^2$$


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