Assume the function $f:\mathbb R \rightarrow \mathbb R$ satisfies the property $f(x+t) \geq f(x)-t^{2}$ Prove that f must be nondecreasing.

Question

Assume the function $f:\mathbb R \rightarrow \mathbb R$ satisfies the property $f(x+t) \geq f(x)-t^{2}$ for all real value of x and all positive value of t. Prove that f must be nondecreasing.

I tried this question with differentiability but is does not work here. Can anyone give some hint?

If it's differentiable this is easy. But I am guessing it need not be. — Matt Samuel, Dec 07 '19 at 15:50

score 1 · Answer 1 · answered Dec 07 '19 at 15:50

1

Hint: You are trying to show $f(x + a) \geq f(x)$ for all $x$ and all $a > 0$. Write the interval $[x,x+a]$ as the union of $N$ intervals of length ${a \over N}$ and use your inequality on each interval. See what you get as a function of $N$.

answered Dec 07 '19 at 15:50

Zarrax

44,950

I choose an interval [x,x+a] and start partition. Let ${x_{1},x_{2},....,x_{n}}$ be the parttion where $x_{j+1}-x_{j}=\frac{a}{n}$. Now $f(x+a)-f(x)=\sum_{j=1}^{n} (f(x_{j+1}-f(x_{j})$ $\geq \sum_{j=1}^{n} -(\frac{a}{n})^{2} = -\frac{{a}^{2}}{n}> 0 $ as $n \rightarrow \infty$ . This implies $f(x+a)>f(x)$. Is this answer is correct? – User124356 Dec 07 '19 at 16:16
Yes, that's correct. To be exact, you take limits as n goes to infinity and get $f(x + a) - f(x) \geq 0$. – Zarrax Dec 07 '19 at 16:44
Yeah. Thanks for the help. – User124356 Dec 07 '19 at 16:46

Assume the function $f:\mathbb R \rightarrow \mathbb R$ satisfies the property $f(x+t) \geq f(x)-t^{2}$ Prove that f must be nondecreasing.

1 Answers1