Can't see what should be a simple $L^p$ inequality

Question

I'm currently reading Brezis Functional Analysis, Sobolev Spaces and PDE's but I'm stuck on what should be an easy inequality. If someone could point out what I'm missing I would greatly appreciate it.

I'm reading the proof that $C_c(\mathbb{R}^N)$ is dense in $L^p(\mathbb{R}^N)$ for $1\leq p<\infty$. Note that $\|\cdot\|_p$ is short hand for $\|\cdot\|_{L^p}$ My inequality is

\begin{equation} \|g-g_1\|_p\leq \|g-g_1\|_1^{\frac{1}{p}} \|g-g_1\|_{\infty}^{1-\frac{1}{p}}\leq \delta^{\frac{1}{p}}(2\|g\|_{\infty})^{1-\frac{1}{p}} \end{equation}

Other relevant information is that $\|g-g_1\|_1\leq \delta$ and $\|g_1\|_\infty \leq \|g\|_\infty$

I can't seem to get from the middle term to the right hand term. My idea was

\begin{equation} \|g-g_1\|_1^{\frac{1}{p}} \|g-g_1\|_{\infty}^{1-\frac{1}{p}}\leq \delta^{\frac{1}{p}} 2^{1-\frac{1}{p}}(\|g\|_{\infty}^{1-\frac{1}{p}}+\|g_1\|_{\infty}^{1-\frac{1}{p}})\leq \delta^{\frac{1}{p}} 2^{1-\frac{1}{p}}2\|g\|_{\infty}^{1-\frac{1}{p}}\end{equation}

but I'm left with an extra $2$. So I imagine I've made a mistake somewhere. many thanks.

Answer 1

I don't see where your additional factor of $2$ is coming from. You have $$\|g - g_1\|_\infty \le \|g\|_\infty + \|g_1\|_\infty \le 2\|g\|_\infty$$ so that $$\|g - g_1\|_\infty^{1 - \frac 1p} \le \left( 2\|g\|_\infty \right)^{1 - \frac 1p}.$$

Answer 2

By the triangle inequality, $$ \lVert g- g_1\rVert_\infty \leq \lVert g\rVert_\infty+\lVert g_1\rVert_\infty \leq 2 \lVert g\rVert_\infty $$ the last inequality from the second assumption. Thus, $$ \lVert g- g_1\rVert_\infty^{1-\frac{1}{p}} \leq (2 \lVert g\rVert_\infty)^{1-\frac{1}{p}} $$ as stated.