# $f(a)=f(b)=f_{+}'(a)=f_{-}'(b)=0$ and $|f''(x)|\le M$. Show that $|f(x)|\le \frac{M(b-a)^2}{16}$.

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 $$f(x_0)\le\frac{M}{16}(b-a)^2$$