Showing that $|f~'(0)|+|f~'(a)| \leq am$.

Question

If $f~''$ is continuous in the interval $[0,a]$ and $|f~''(x)| \leq m ~\forall~x \in [0,a].$ Assume that $f$ takes on it's largest value at an interior point of this interval. Show that $|f~'(0)|+|f~'(a)| \leq am$.

Attempt: Since, $f$ takes it's largest value at an interior point $c \in [a,b]$

$ \implies f~'(c)=0 , f~''(c)<0$

The least possible negative value of $f~''(c)=-m$

How do I move ahead with this problem?

Thank you for your help.

EDIT: Thanks for the answers. I have one query though. What could be the use of specifying the condition that $f~''(x)$ is continuous throughout the given interval?

score 5 · Accepted Answer · answered Jan 18 '15 at 15:52

5

Hint

$$f'(0)=f'(c)+f''(\xi_{1})(0-c)=-cf''(\xi_{1})$$ $$f'(a)=f'(c)+f''(\xi_{2})(a-c)=(a-c)f''(\xi_{2})$$ so $$|f'(0)|+|f'(a)|=c|f''(\xi_{1})|+(a-c)|f''(\xi_{2})|\le cm+(a-c)m=am$$

answered Jan 18 '15 at 15:52

math110

93,304

Thank you for the answer :) – MathMan Jan 18 '15 at 16:43
I have one query though. What could be the use of specifying that $f~''(x)$ is continuous? Thanks – MathMan Jan 18 '15 at 17:37
@Wanderer: In this proof you need $f''$ to be continuous, for MVT. For my proof also you need it for Fundamental theorem of Calculus to hold. – voldemort Jan 18 '15 at 20:01

score 3 · Answer 2 · answered Jan 18 '15 at 15:53

3

Let $c$ be the interior point with $f'(c)=0$.

Now, $|\int\limits_{0}^cf''|=|f'(c)-f'(0)|=|f'(0)|$.

On the other hand, $|\int\limits_{0}^cf''| \leq \int\limits_{0}^c|f''|\leq mc$.

Now apply the same trict to $\int\limits_{c}^a f''$

answered Jan 18 '15 at 15:53

voldemort

13,182

Showing that $|f~'(0)|+|f~'(a)| \leq am$.

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