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Is this a proof? The problem states: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $\int_{x}^{1}f(t)dt\geq(1-x)^{2}$, for any $x\in\mathbb{R}$. Prove that $f(1)=0$.

I denoted by $F(x)=\int_{x}^{1}f(t)dt$ and I noticed that $F(1)=0$ and $F(x)\geq0=F(1)$ and now, $f(1)=\lim_{x\rightarrow1,x>1}\frac{F(x)-F(1)}{x-1}\geq0$ and $f(1)=\lim_{x\rightarrow1,x<1}\frac{F(x)-F(1)}{x-1}\leq0$, so $f(1)=0$. Is this correct?

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    $\begingroup$ Looks fine to me. $\endgroup$ – Kavi Rama Murthy Mar 6 at 5:27
  • $\begingroup$ Well, $F(x)\geqF(1)$ and from there result the inequalities. $\endgroup$ – Septimiu Cristian Mar 6 at 7:04
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This is quite elementary

for $x\leq1$ we have

the area under $f(x) $ is $\geq0$ since $(1-x)^2\geq0$

$\implies f(t)\geq0 $ for $x<1$

for $x\geq1$ we switch our limits giving us a negative negative sign our equation becomes as follows

$-\int_{1}^{x}f(t)dt\geq(1-x)^{2}$,

multiplying with a negative sign we get

$\int_{1}^{x}f(t)dt\leq-(1-x)^{2}$,

the area under $f(x)$ is $\leq0$ since $-(1-x)^2\leq0$

$\implies f(t)\leq0$ for $ x>1$

using intermediate value theorem since $f(x)$ is greter than 0 for $x<1$ and less than 0 for $x>1$

$\implies f(1)=0$

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