I want to show that $\| \cdot \|_p$ is equivalent to $\| \cdot \|_q$ iff $p=q$ on $\mathbb{R}$.   I think the best way to do this is via a contradiction. Supposing that $\| \cdot \|_p \leq \alpha \| \cdot \|_q \leq \beta \| \cdot \|_p$, we have $$\left( \int_{\mathbb{R}} \left| f \right|^p d\mu \right)^{1/p} \leq \alpha \left( \int_{\mathbb{R}} \left| f \right|^q d\mu \right)^{1/q} \leq \beta \left( \int_{\mathbb{R}} \left| f \right|^p d\mu \right)^{1/p}.$$ I'm then thinking of applying Holder's inequality, but haven't been successful .
121151
Are all the norms of $L^p$ space equivalent? That is, for any $p,~q \in R^+$, there exist two positive number $C_1,~C_2$ such that $$ C_1\|u\|_{L^q} \le \|u\|_{L^p} \le C_2\|u\|_{L^q}. $$
86181
Let $(X,\mathcal{M},\mu)$ be a measurable space, $1&lt;p_{k}&lt;\infty$ for $k=1,2,\ldots,N$ with $\sum_{k=1}^{N}\frac{1}{p_{k}}=1$ and $f_{k}\in L^{p_{k}}(\mu)$ for $k=1,\ldots,N$.  Show that $$\left\|f_{1}f_{2}\cdots f_{N}\right\|_{1}\leq \|f_{1}\|_{p_{1}}\|f_{2}\|_{p_{2}}\cdots\|f_{N}\|_{p_{N}}.$$  Remark: I tried to use the Holder inequality with $p=p_{1} $ and $q=\frac{1}{\sum_{k=2}^{\infty}\frac{1}{p_{k}}}$, but when I try to continue the process, it becomes complicated until I think that this is not the way.
53609
Let $f(x) = \frac{1}{2}x^tQx-b^tx, Q\in\mathbb{R}^{n\times n}$   symmetric definite positive and $b\in\mathbb{R}^n$. Let $x^0, x^1,  \cdots, x^n\in\mathbb{R}^n$ and define $\delta^j = x^j-x^0$, $\gamma^j  = \nabla f(x^j)-\nabla f(x^0)$, $j=0,1\cdots,n$. Prove that the vectors $\{\delta^j\}_{j=1}^n$ are linearly independent, then   $$\overline{x} = x^n-[\delta^1\cdots \delta^n]\cdot[\gamma^1\cdots  \gamma^n]^{-1}\cdot\nabla f(x^n)$$ is a global minimizer of $f$   So, that's what I did:  $$\nabla f(x) = Qx-b^t\implies \gamma^j = Qx^j-b^t-(Qx^0-b^t) = Q(x^j-x^0)\implies\\\gamma^j = Q\delta^j\implies\delta^j = Q^{-1}\gamma^j$$  So  $$\overline{x} = x^n-[Q^{-1}\gamma^1\cdots Q^{-1}\gamma^n]\cdot[\gamma^1\cdots  \gamma^n]^{-1}\cdot\nabla f(x^n) =\\ x^n -Q^{-1}[\gamma^1\cdots \gamma^n][\gamma^1\cdots \gamma^n]^{-1}\cdot\nabla f(x^n) = x^n-Q^{-1}\nabla f(x^n) \implies \\\overline{x} = x^n-Q^{-1}(Qx^n-b^t) = x^n-x^n+Q^{-1}b^t = Q^{-1}b^t$$  So I need to prove $\overline{x} = Q^{-1}b^t$ is a global minimizer? Also, I need to use linear independence of $\{\delta^j\}$ to something. I guess it is related with the inverse matrix. Maybe linear independence of $\{\delta\}^j$ implies that that matrix has an inverse, but I don't know why.  Also, as I recall, $f(x)$ is quadratic and therefore convex. So shouldn't it has only one minimum? (therefore, since it is convex, it's the global?) So I should need to prove only that $\overline{x}$ is a local minimum? In that case I should only have to calculate $\nabla f(\overline{x}) = Q(Q^{-1}b^t)-b^t = 0$. The second derivative is just $Q$ and is definite positive. So: first derivative zero, second derivatie def. positive. Therefore this point $\overline{x}$ should be a minimum.  Even though I think I solved the exercise I don't know what all this means. It could just have asked if $Q^{-1}b^t$ is a global minimum. I think it is trying to show me a process of finding a global minimum for a quadratic function but I don't see the motivation.
15088
If $f \in L_p \cap L_q$ and $1 \leq p &lt; r &lt; q &lt; \infty$ then $f \in L_r$.   Proof: Let $E = \{ x \in X: \left\lvert f \right\rvert \leq 1 \}$. Then, $$\int \left\lvert f \right\rvert^r \mathrm{d}\mu = \int_E \left\lvert f \right\rvert^r \mathrm{d}\mu+\int_{E^C} \left\lvert f \right\rvert^r \mathrm{d}\mu \leq \int_E \left\lvert f \right\rvert^p \mathrm{d}\mu+\int_{E^C} \left\lvert f \right\rvert^q \mathrm{d}\mu$$ $$\leq \int \left\lvert f \right\rvert^p \mathrm{d}\mu+\int \left\lvert f \right\rvert^q \mathrm{d}\mu &lt; \infty.$$ Thus, $f \in L_r$.  I just want to make sure I am somehow not missing anything. Thanks
68060
I want to show that $$f: \mathbb{R}_{\geq0} \to \mathbb{R}_{\geq0}$$ $$x \mapsto x^s$$ is Holder continuous with Holder exponent $s \in \mathbb{R}$, where $0&lt;s \leq 1$. So what I want to show is that $\exists \hspace{2 mm} C \in \mathbb{R}_{\geq0}$ sucht that for all $ x,y \in \mathbb{R}_{\geq0},  $ $$|x^s -y^s| \leq C|x-y|^s$$ and therefore, assuming, wlog $\hspace{1mm} x&gt;y$ $$(x^s -y^s) \leq C(x-y)^s.$$  I thought about Bernoulli's inequality but couldn't make that work.I thought about the binomial theorem, but didn't know how to handle the fact that $s \in \mathbb{R}$.
93071
Let $p,q,r \in (1,\infty)$ with $1/p+1/q+1/r=1$. Prove that for every functions   $f \in L^p(\mathbb{R})$, $g \in L^q(\mathbb{R})$,and $h \in L^r(\mathbb{R})$    $$\int_{\mathbb{R}} |fgh|\leq \|f\|_p\centerdot \|g\|_q \centerdot\|h\|_r.$$   I was going to use Hölder's inequality by letting  $1/p+1/q= 1/(pq/p+q)$ and WLOG  let $p&lt;q$ so that $L_q(\mathbb{R})\subseteq L_p(\mathbb{R})$, but I cannot use this inclusion because $\mathbb{R}$ does not have finite measure.   Would you please help me if you have any other method to approach this problem?
31280
My question is:  Let $ V $ be a vector space (over $ \mathbb K\in\{\mathbb{R}, \mathbb{C}\} $), $ X,Y\subseteq V $ two subspaces equipped with norms $ \|\cdot\|_X, \|\cdot\|_Y $ such that $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$ are Banach spaces and $ D\subseteq X\cap Y$. If $ D $ is dense in $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$, is $ D $ also dense in $ X\cap Y $ equipped with $ \|\cdot\|:=\|\cdot\|_X + \|\cdot\|_Y $?  At first sight, it seemed very clear to me that this should be true. But I even fail to answer the following (possibly) easier question:  Let $ X $ be a vector space over $ \mathbb K $ equipped with two norms $ \|\cdot\|_1, \|\cdot\|_2 $ such that $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_1)$ are Banach spaces and $ D\subseteq X$. If $ D $ is dense in $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_2)$, is $ D $ also dense in $ X $ equipped with $ \|\cdot\|:=\|\cdot\|_1 + \|\cdot\|_2 $?  The answer is yes, if $ \|\cdot\|_1, \|\cdot\|_2 $ are equivalent, so I tried to think about a counterexample using nonequivalent norms on a specific space and I also found a nice paper about nonisomorphic complete norms () but it didn't helped me so far to construct anything useful for my question.  Thanks for your help!
80732
Trying to find a proof for Clarkson's inequality, which states that if $2 \leq p &lt; \infty$, then for any $f, g \in L^p$, we have that $$\left|\left|f+g\right|\right|_p^p + \left|\left|f-g\right|\right|_p^p \leq 2^{p-1}\left(||f||_p^p + ||g||_p^p\right) $$ I have a couple of ideas, but I know they are wrong. I tried saying that $|f+g|^p \leq |f|^p + |g|^p$ and the same thing for $|f-g|$ but I know those inequalities aren't true in general.
31686
Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated.
87077
Suppose $\mu(X)&lt;\infty$. Show that if $\ f\in\ L^q(X,\mu)$ for some $1\le q&lt;\infty$ then $\ f\in\ L^r(X,\mu)$ for $1\le r&lt;q$   In the solution the assistant wrote this:  If $1\le r&lt;q$ then $q/r\ge 1$. Let p be the dual exponent to $q/r$ (i.e. $\frac{1}{q/r}+\frac{1}{p}=1)$  then by Hölder Inequality  $\int_{X}|f|^rd\mu\le\underbrace{(\int_{X}|f|^qd\mu)^{r/q}}_{??????}(\int_X1^pd\mu)^{1/p}&lt;\infty$  for me the middle part doesn't make sense (it is not the hölder inequality) can somebody show it ?  instead of this can't we say (only in this case) that   $\int_{X}|f|^rd\mu\le (\int_{X}|f|d\mu)^r$ then  $(\int_{X}|f|d\mu)^r\le ((\int_{X}|f|^qd\mu)^{1/q}(\int_X1^pd\mu)^{1/p})^r&lt;\infty$  ?
101605
I am trying to prove the above statement but I'm not sure if my proof is correct. My proof is as follows,  Given $u\cdot v$, we know by the C-E Inequality that $|u \cdot v| \leq ||u|| \ ||v||$  Consider $\frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2$  We can manipulate this as follows $$\frac{1}{4}(||u+v||^2 - ||u-v||^2) = \frac{1}{4}(||u+v||- ||u-v||)(||u+v||+||u-v||)$$ Using the triangle inequality the above can be written as follows, $$...\leq \frac{1}{4}(||u|| +||v|| -||u||+||v||) (||u|| +||v|| +||u||-||v||) = \frac{1}{4} (4||u||||v||) = ||u||||v||$$  As such both $u\cdot v$ and $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ are $\leq ||u|| ||v||$  I'm not sure whether I've proven this correctly because I ended up proving that both sides of the equation are less than or equal to something entirely different (the magnitude of v multiplied by the magnitude of u). Is this acceptable? I don't think it is, since if both can be less than or equal to the magnitudes multiplied then they do not necessarily have to be equal to each other. If this is the case, can someone give me a hint on how to move forward with this proof?
114411
How to prove that $\ell^p \setminus \cup_{1\leq q &lt;p}\ell^q$ is not empty?  Here $\ell^p=L^p(\mathbb{N})$.
4583
$\def\d{\mathrm{d}}$Given that $X_1, X_2, \cdots$ are i.i.d., prove that$$ \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \xrightarrow{\mathrm{a.s.}} 0 \Longleftrightarrow E(|X_1|) &lt; +\infty. $$   My try:  Because $\varliminf\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ and $\varlimsup\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ are both measurable with respect to the exchangeable σ-algebra of $X_1, X_2, \cdots$, they are both almost surely constants by the Hewitt-Savage 0-1 law, i.e.$$ \varliminf_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_1, \quad \text{a.s.} \tag{1}\\ \varlimsup_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_2, \quad \text{a.s.} $$ where $c_1, c_2 \in [0, +\infty]$. Also, if the distribution function of $|X_1|$ is $F(x)$, then the distribution function of $\dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ is$$ F_n(x) = P\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \leqslant x \right) = \prod_{k = 1}^n P(|x_k| \leqslant nx) = (F(nx))^n, $$ which implies\begin{align*} E\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \right) &amp;= \int_0^{+\infty} x \,\d((F(nx))^n) = \frac{1}{n} \int_0^{+\infty} y \,\d((F(y))^n)\\ &amp;= \frac{1}{n} \int_0^{+\infty} y·n (F(y))^{n - 1} \,\d F(y) = \int_0^{+\infty} y (F(y))^{n - 1} \,\d F(y)\\ &amp;\leqslant \int_0^{+\infty} y \,\d F(y) = E(|X_1|). \end{align*} Also for any $c &gt; 0$,$$ \sum_{k = 1}^\infty P(|X_1| \geqslant ck) &lt; \infty \Longleftrightarrow E(|X_1|) &lt; +\infty. \tag{2} $$ However, I failed to connect (1) and (2) to prove the proposition.  Any ideas are appreciated. Thanks in advance.
69557
Suppose we have two equivalent norms, $\|\cdot \| \sim \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert$. We assume that $x_n \rightarrow x $ in $(X,\|\cdot \|)$. Show that $x_n \rightarrow x $ in $(X,\left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert)$  We observe since the norms are equivalent, there exits constants $\alpha, \beta$, such that $\alpha \|\cdot \| \leq \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert \leq \beta \|\cdot \|$, with $0 \lt \alpha \leq \beta$  Since, $x_n \rightarrow x$ in $(X,\|\cdot \|$) we know that for all $\epsilon&gt;0, \exists K(\epsilon)$ such that for n>$K(\epsilon),$ we have $\|x_n -x\| \leq \epsilon$  Since, we know $\beta \gt 0$ , then $ \beta \|x_n -x\| \leq \beta \epsilon$, Thus we see that $\left\vert\!\left\vert\!\left\vert x_n-x\right\vert\!\right\vert\!\right\vert\leq \beta \epsilon$.  This is how far I've gotten, our goal to is to find a a $J(\epsilon)$ such that for all $\epsilon&gt;0, \exists J(\epsilon)$ such that for n>$J(\epsilon),$ we have $\left\vert\!\left\vert\!\left\vert x_n-x \right\vert\!\right\vert\!\right\vert \leq \epsilon$  My candidate for $J(\epsilon$) is $J(\epsilon) = \frac{K(\epsilon)}{\beta}$ from my above argument, but I'm not exactly sure how to validate this. If someone could help me on the proof, it would be very appreciated. I believe I'm pretty close, just missing something small.
24201
The following is a well-known result in functional analysis:     If the vector space $X$ is finite dimensional, all norms are equivalent.     Here is the standard proof in one textbook. First, pick a norm for $X$, say  $$\|x\|_1=\sum_{i=1}^n|\alpha_i|$$ where $x=\sum_{i=1}^n\alpha_ix_i$, and $(x_i)_{i=1}^n$ is a basis for $X$. Then show that every norm for $X$ is equivalent to $\|\cdot\|_1$, i.e.,  $$c\|x\|\leq\|x\|_1\leq C\|x\|.$$ For the first inequality, one can easily get $c$ by triangle inequality for the norm. For the second inequality, instead of constructing $C$, the  is applied to construct a contradiction.   The strategies for proving these two inequalities are so different. Here is my question,      Can one prove this theorem without Bolzano-Weierstrass theorem?   UPDATE:     Is the converse of the theorem true? In other words, if all norms for a vector space $X$ are equivalent, then can one conclude that $X$ is of finite dimension?
87446
Let $x\in\mathbb{C}^n$. Do the following inequalities hold? $$\lVert x\lVert_2\leq\sqrt{\lVert x\lVert_1\lVert x\lVert_\infty}\leq\frac{1+\sqrt{n}}{2}\lVert x\lVert_2.$$ I think the first inequality is the Holder inequality with $p=1$ and $q=\infty$.
21585
I need help proving or disproving this! If anyone could show me the sequence of steps I can use to help me show the left is equal to the right I would be grateful.  $$\mathcal P(A-B)=\mathcal P(A)-\mathcal P(B)$$