Let $f:\mathbb{R}\rightarrow [0,\infty)$ be a continuous function. Then which of the following is not true?

Question

(a) There exist a $x\in \mathbb{R}$ such that $f(x)=\frac{f(0)+f(1)}{2}$

(b) There exist a $x\in \mathbb{R}$ such that $f(x)=\sqrt{f(-1)f(1)}$

(c) There exist a $x\in \mathbb{R}$ such that $f(x)=\int_{-1}^{1}f(t)dt$

(d) There exist a $x\in \mathbb{R}$ such that $f(x)=\int_{0}^{1}f(t)dt$

Attempt

For $(a)$ take $f(x)=5$, for $(b)$ take $f(x)=1$. So a and b can be false in general.

I need some hints for $c$ and $d$. Can I use fundamental theorem of Calculus here?

a) and b) are true and you are proving them using an example. You cannot prove a result with just one example. — Kavi Rama Murthy, Commented Oct 2, 2018 at 10:33
@KaviRamaMurthy yes I know that! I was trying to reject some option using some examples as I have to find the condition on $f$ which can not be met. — Shweta Aggrawal, Commented Oct 2, 2018 at 10:34

José Carlos Santos · Accepted Answer · 2018-10-02 10:38:27Z

6

Assertion (c) is false in general. Just take $f(x)=1$. All the others are a consequence of the intermediate value theorem.

answered Oct 2, 2018 at 10:38

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$\begingroup$ Can you please give some hint how intermediate value theorem will be used for $d$ part. I have proved that statement for $a$ and $b$ using IVT $\endgroup$
– Shweta Aggrawal
Commented Oct 2, 2018 at 10:48
3

$\begingroup$ For every function continuous function $f\colon[0,1]\longrightarrow\mathbb R$, if $m$ is its minimum and $M$ is its maximum, then$$m\leqslant\int_0^1f(x)\,\mathrm dx\leqslant M.$$ $\endgroup$
– José Carlos Santos
Commented Oct 2, 2018 at 10:59
1

$\begingroup$ Since restriction of continuous function is continuous, consider $f$ restricted to $[0,1]$. Now a continuous function on closed and bounded interval attain its bounds. Since $f$ is continuous it is Riemann Integrable. Now we know $\int f$ is bounded by $m(b-a)$ from left and $M(b-a)$ from right. In our case $b-a=1$. By IVT $f$ takes every value between $m$ and $M$ and we are done.!! Is this correct? $\endgroup$
– Shweta Aggrawal
Commented Oct 2, 2018 at 11:05
2

$\begingroup$ Completely correct! $\endgroup$
– José Carlos Santos
Commented Oct 2, 2018 at 11:07
$\begingroup$ Thanks for discussion. I learned some new mathematics today. Also I learned how to apply the things I already know like IVT. $\endgroup$
– Shweta Aggrawal
Commented Oct 2, 2018 at 11:08

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