First I start with the definition: a function $f:\mathbb{N} \rightarrow \mathbb{R}^{+}$ is b-smooth for an integer $b \geq 2$ if $f$ is eventually non decreasing and if $$ \exists c \in \mathbb{R}^{+} \exists n_0 \in \mathbb{N} \forall n \geq n_0 \hspace{1cm} f(b*n) \leq c*f(n)$$
The function is smooth if the b-smoothness holds for all integers $b \geq 2$.
Now I need to prove that $f(n)=n*log(n)$ is b-smooth for all $b \geq 2$.
I started like this
$f(b*n)=bn*log(bn) \leq c*n*log(n)$
$bn*log(bn) \leq c*n*log(n)$
I could divide both sides by n then $b*log(bn) \leq c*log(n)$ now
$\frac{log(bn)}{log(n)} \leq \frac{b}{c}$ with the rules of logarithm I can write it as $\frac{log(bn)}{log(n)}=log_{n}(bn) \leq \frac{b}{c}$
But how can I show that the function is bounded by $\frac{b}{c}$ ? Maybe my steps were completely wrong
Just take $n_0=b$ and $c=2b$. Then, since $\log$ is increasing: $$ \implies \log(b)\leq \log(n) $$ $$ \implies bn\log(b)\leq bn\log(n) $$ $$ \implies bn\log(b)+bn\log(n)\leq bn\log(n)+bn\log(n) $$ $$ \implies bn\log(bn)\leq 2bn\log(n) $$