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

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

(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. 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. Commented Oct 2, 2018 at 10:34
• Your counter example does not contradict with a) and b). Commented Jul 10, 2019 at 13:40

Assertion (c) is false in general. Just take $$f(x)=1$$. All the others are a consequence of the intermediate value theorem.
• 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 Commented Oct 2, 2018 at 10:48
• For every function continuous function $f\colon[0,1]\longrightarrow\mathbb R$, if $m$ is its minimum and $M$ is its maximum, then$$m\leqslant\int_0^1f(x)\,\mathrm dx\leqslant M.$$ Commented Oct 2, 2018 at 10:59
• 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? Commented Oct 2, 2018 at 11:05