Now for u′(x)≤0 to be true for some x, u"(x)<0 for some x∈[0,∞) as u′(0)>0. How you concluded this line?

@DevanshBhardwaj

For u''(x)<=0 implies u'(x) is strictly decreasing and in an alter way u'(x) not strictly decreasing implies u''(x) is not <=0.

@AasthaChoudhary $\int_0^xu"(x)\,dx=u'(x)-u(0)$. So u" must have been negative somewhere for $u'(x)-u(0)<0$ or $u'(x)<u(0)$.

sorry u'(x)-u'(0)

Give me some time.I am not getting things clearly.