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How would I formally prove that $|x\log x| \le 1$ on $(0,1]$. I tried showing that $x\log x \le x^2 \le 1$, but even showing $\log x \le x$ I'm finding difficult. And then finding a lower bound for the absolute value is driving me nuts. My apologies if this is trivial, but I just can't quite get it. I'm not allowed to use calculus techniques such as second derivative tests and such. Any aid would be great.

Anirban Niloy
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  • I would be surprised if such a no calculus solution exists, because I'm not sure how you can even define log(x) without some calculus, and so I'm not sure what properties of log you'd be allowed to use without 'calculus techniques.' –  Feb 28 '19 at 15:12
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    I hesitate to post this as an answer, because like the previous commenter I cannot see how "calculus techniques" in some sense can be altogether avoided. If one is allowed to use the definition of $\log x$ as an integral, then $$|x\log x| = -x\log x = x\int_x^1\frac{dt}{t} = \int_x^1\frac{x}{t},dt \leqslant 1 - x.$$ I'll get me coat. :) – Calum Gilhooley Feb 28 '19 at 16:04
  • How would one show the first equality. –  Feb 28 '19 at 17:05

2 Answers2

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Since $x\log x\le0$ for $0<x\le1$, you want to prove that $x\log x\ge-1$, which means $$ \log x\ge-\frac{1}{x} $$ that's the same as $$ \log\frac{1}{x}=-\log x\le \frac{1}{x} $$ With the substitution $t=1/x$, this becomes proving that, for $t\ge1$, $\log t\le t$, or $t\le e^t$.

Now choose the proof you prefer among those at Simplest or nicest proof that $1+x \le e^x$

With a bit of calculus, consider the function $f(x)=x\log x$; then $$ f'(x)=1+\log x $$ which is negative for $0<x<1/e$ and positive for $x>1/e$.

Since $\lim_{x\to0}f(x)=0$, $f(1/e)=-1/e$ and $f(1)=0$ you can say even more: $$ -\frac{1}{e}\le x\log x\le 0 $$ for $x\in(0,1]$.

egreg
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$\log(x) \leq x$ is trivial, but does not help you because $\log(x)$ is negative in $(0,1]$.

For this solution, I assume that you know $y+1 \leq \exp(y)$ for $y \geq 0$, $-\log(x) = \log\left(\frac{1}{x}\right)$ for $x \in (0,1]$ and that the logarithm is monotone. All of these facts can be proven directly by the definition of $\exp$ and $\log$ (as the inverse of $\exp$).

Let $y \geq 0$ and observe that $y+1 \leq \exp(y)$ implies $\log(y+1) \leq y$ by monotonicity. Plugging in $y = \frac{1}{x}-1$, we get $$-\log(x) = \log\left(\frac{1}{x}\right) \leq \frac{1}{x}-1$$ and hence $|x\log(x)| = -x\log(x) \leq 1 - x \leq 1$ for $x \in (0,1]$.

Klaus
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