### CSIR JUNE 2011 PART C QUESTION 68 SOLUTION ($||x||_1 \le d \,||x||_{\infty}$)

For $(x_1,x_2,\dots,x_d) \in \Bbb R^d$, and $p \ge 1$ define $$||x||_p = (\sum\limits_{i=1}^d|x_i|^p)^{\frac{1}{p}}$$ and $$||x||_{\infty} = \text{max }\{x_i : 1 \le i \le d\}.$$ Which of the following inequalities hold for all $x \in \Bbb R^d$?
1)$||x||_1 \ge ||x||_2 \ge ||x||_{\infty}$,
2)$||x||_1 \le d \,||x||_{\infty}$,
3)$||x||_1 \le \sqrt d \,||x||_{\infty}$,
4)$||x||_1 \le \sqrt d \,||x||_2$.
Solution: Assume $p \ge 1$.

option 1: (True)We have, for $1 \le i \le d$, $|x_i|^p \le (||x||_p)^p.$ Hence $$||x||_{\infty} \le ||x||_p \,\,\,\,\,\,\,\,\,\,(\text{ for any }p \ge 1).$$
ClaimIf $p \le q$ then $||x||_q \le ||x||_p$.
Proof:
Assume $0<a\le 1$, then $$(\sum_{i=1}^d |x_i|)^a \le \sum_{i=1}^n |x_i|^a.$$ Because, $\frac{\sum_{i=1}^d |x_i|^a}{(\sum_{j=1}^d |x_j|)^d} = \sum_{i=1}^d \frac{|x_i|^a}{(\sum_{j=1}^d |x_j|)^a} = \sum_{i=1}^d (\frac{|x_i|}{\sum_{j=1}^d |x_j|})^a \ge \sum_{i=1}^d \frac{|x_i|}{\sum_{j=1}^d|x_j|} = 1$.
Note that, in the last step, we have used the fact that $x^a > x$ if $x, a \in (0,1]$.
Now, assume $p \le q$ then $\frac{p}{q} \le 1$. By using the above result, We have $$||x||_q = \big(\sum_{i=1}^d|x_i|^q\big)^{\frac{1}{q}} = \big(\big(\sum_{i=1}^d |x_i|^q\big)^{\frac{p}{q}}\big)^{\frac{1}{p}} \le (\sum_{i=1}^d (|x_i|^{q})^{\frac{p}{q}})^{\frac{1}{p}} = \sum_{i=1}^d |x_i|^p)^{\frac{1}{p}} \\ = ||x||_p.$$
In particular we have, $$||x||_2 \le ||x_1||.$$
option 4: (True)The Cauchy-Schwarz inequality in $\Bbb R^d$:
Let $x = (x_1,x_2,\dots,x_d), y=(y_1,y_2,\dots,y_d)$. Then
$$(\sum_{i=1}^d x_iy_i)^2 \le (\sum_{i=1}^d x_i^2)(\sum_{i=1}^d y_i^2).$$ Substitute $x = (1,1,\dots,1) \in \Bbb R^d$ in this equation we get the required result.
option 2:(True) Let $x=(x_1,x_2,\dots,x_d) \in \Bbb R^d$ and $a = \text{max }_{1 \le i \le n}|x_i| = ||x||_{\infty}$. First, we claim that $||x||_2 \le \sqrt d \,||x||_{\infty}$. We have $(||x||_2)^2 = \sum_{i=1}^d |x_i|^2 \le \sum_{i=1}^d a^2 = (||x||_{\infty})^2 (\sum_{i=1}^d 1) = d (||x||_{\infty})^2.$
This proves the claim.
Now, using this result in option 4 we get $$||x||_1 \le \sqrt d\, ||x||_{2} \le d \,||x||_{\infty}.$$
option 3: (False) Take $x = (1,1) \in \Bbb R^2$ and calculate the LHS and RHS. We get $2 \le \sqrt 2$.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
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