Prove that $||(\sum^k_{i=1}|f_i|)^{\frac{1}{2}}||_p \leq (\sum^k_i ||f_i||^2_p)^{\frac{1}{2}}$

Question

In one popular Chinese Real Analysis textbook, I met a problem stated as the title: Prove the following inequality \begin{equation} \left|\left|\left(\sum^k_{i=1}|f_i|\right)^{\frac{1}{2}}\right|\right|_p \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}}, \end{equation} given the condition: $2\leq p < \infty, f_i \in L^p(E) (i=1,2,...,k)$.

Here is the snapshot .

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My attempts are as follows:

1. It suffices to consider $k=2$.
2. Without loss of generality, assume all $f_i>0$.
3. Put the $\frac{1}{2}$ on LHS out of the norm, and cancel it with the $\frac{1}{2}$ on RHS.
4. Put the $2$ on RHS into the p-norm. It becomes:

\begin{equation} ||\sum^k_{i=1}f_i ||_{\frac{p}{2}} \leq \sum^k_i \left|\left| f_i^2 \right|\right|_{\frac{p}{2}}, \end{equation}

The condition $2\leq p < \infty$ reminds me the Clarkson inequalities, and in fact they appeared slightly before this problem. Yet the book didn't provide any hint.

Then I'm stucked. Can anyone help me?

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Thank you for the comments!

Though the book has rare typos, the original inequality may have square $2$ outside the LHS p-norm? That is,

\begin{equation} \left|\left|\left(\sum^k_{i=1}|f_i|\right)^{\frac{1}{2}}\right|\right|_p^2 \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}}, \end{equation}

such that the simplified one is

\begin{equation} ||\sum^k_{i=1}f_i ||_{\frac{p}{2}}^2 \leq \sum^k_i \left|\left| f_i^2 \right|\right|_{\frac{p}{2}}, \end{equation}

The reason is that, similar to Cauchy, Minkowski, Holder inequalities, both sides now have the same "order"/"power".

No matther which case is true, this kind of inequality is elegant. Hope we can achieve the right one.

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@Calvin Khor I think your comment is correct. In this way, the question is to show:

\begin{equation} \left|\left|\left(\sum^k_{i=1}|f_i|^2\right)^{\frac{1}{2}}\right|\right|_p \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}}, \end{equation}

Then the proof is obvious!

Answer 1

Following the comment of @Calvin Khor, and based on the difficulty of the problem it should be in this book, here I prove the following statement:

\begin{equation} \left|\left|\left(\sum^k_{i=1}|f_i|^2\right)^{\frac{1}{2}}\right|\right|_p \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}}, \end{equation}

given the condition that $2\leq p < \infty, f_i \in L^p(E) (i=1,2,...,k)$.

Proof: It suffices to show

\begin{equation} \left|\left|\left(\sum^k_{i=1}|f_i|^2\right)\right|\right|^{\frac{1}{2}}_{\frac{p}{2}} \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}}, \end{equation}

\begin{equation} \left|\left|\left(\sum^k_{i=1}|f_i|^2\right)\right|\right|_{\frac{p}{2}} \leq \sum^k_i \left|\left|f_i\right|\right|^2_p, \end{equation}

\begin{equation} \left|\left|\left(\sum^k_{i=1}|f_i|^2\right)\right|\right|_{\frac{p}{2}} \leq \sum^k_i \left|\left| |f_i|^2\right|\right|_{\frac{p}{2}}, \end{equation}

where the last inequality holds by Minkowski inequality. This completes the proof.