Let $f:\mathbb{R}^n\rightarrow ]-\infty,+\infty]$ be proper and strictly convex. Show that $f$ has at most one minimizer.  I am hoping someone can give me some feedback on the proof for this. I feel that it may not be rigorous enough.  proof: Let $\mu=\inf_{x\in\mathbb{R}^n} f(x)$ (i.e. $\mu$ is a global infimum) and choose a sequence $(x_n)_{n=1}^{\infty}$ in $\mathbb{R}^n$ such that $f(x_n)\rightarrow\mu.$ Then $(x_n)$ is a bounded sequence   [otherwise $(x_n)\rightarrow+\infty$, so $f(x_n)\rightarrow+\infty$. Therefore, $\mu\rightarrow+\infty$, which means that $dom(f)=\emptyset$, which negates our assumption that $f$ is proper.]  Moreover, by the Bolzano-Wierstrass theorem there exists a convergent subsequence $(x_{n_{k}})_{k=1}^{\infty}$, which means $(x_{n_{k}})\rightarrow x\in\mathbb{R}^{n}$. Thus, we see that \begin{align} \mu&amp;\leq f(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\qquad\text{as $f$ is lsc}\\ &amp;\leq\liminf_{k \to \infty} \;(f(x_{n_k}))\\ &amp;=\mu. \end{align}  Thus, $\mu=f(x)$ is our minimizer, and since $f$ is strictly convex we get that $\mu$ is unique.$\qquad\;\;\;\;\Box$
123458
Let $f\colon X \to [ -\infty, +\infty]$ be proper and strictly convex. Show that $f$ has at most one minimizer. Give examples where $f$ is strictly convex, and either (i) $f$ has one minimizer; or (ii) $f$ has no minimizer.  For the strict convexity part, you may use the fact that f is strictly convex whenever its Hessian is positive definite.   I am not sure how to show that f has at most one minimizer, could someone help get me started with the question?
20383
Let $A \subset \mathbb{R}^n$ be closed and $d_A(x) = \inf_{a \in A} \|x-a\|$. I need to show that $d_A$ is convex if and only if $A$ is convex.  I have no good ideas in either direction. Can somebody please give me a hint?  For the direction ($\Longleftarrow$), my attempt was:  Let $a_1 := \operatorname*{argmin}_{a \in A} \|x-a\|, a_2 := \operatorname*{argmin}_{a\in A} \|y-a\|$, then:  \begin{align*} d_A(x) &amp;= \inf_{a \in A} \|\lambda x + (1 - \lambda)y - a\| = \inf \|\lambda(x-a) + (1-\lambda) (y-a)\|\\ &amp;\leq \lambda \|x-a_1\| + (1-\lambda)\|y-a_2\| = \lambda d_A(x) + (1-\lambda)d_A(y) \end{align*}  However, I'm not sure whether the inequality in the third step is actually correct.
54751
Let $f\in C^2(\mathbb{R},\mathbb{R})$.   Assume there exist  $(a,b) \in \mathbb{R}^2$ such that  $\forall x \in \mathbb{R}, (f(x))^{2} \leqslant a$ and $(f'(x))^{2} + (f''(x))^{2} \leqslant b$.   Prove that: $$(f(x))^{2} + (f'(x))^{2} \leqslant \max(a,b)$$ for all $x \in \mathbb{R}$.   I tried to use taylor lagrange with the integral form of the remainder but I did not succeed.  Ì really don't know how to tackle  the problem.   Any suggestion, Hint (or Answer) will be very helpful,  Thank you in advance,
71449
Let $f:\mathbb{R}^n\to \mathbb{R}$ be a convex function. It is well-known that (see e.g.~) that there exists a set $A$ such that  $$ f(x)=\sup_{a\in A}A(a)x+b(a), \quad \forall x\in \mathbb{R}^n, $$ where $A(a)\in\mathbb{R}^{n\times n}$ and $b(a)\in \mathbb{R}^n$.  I was wondering, is there any reference saying we can choose $A$ as a countable set?
109086
Show that the function $f: S \to \mathbb R$ given by $$f(x,s,t):=-\ln(st - ||x||^2)$$ is convex on $$S := \left\{(x,s,t) \in \mathbb R^n \times \mathbb  R \times \mathbb R: \frac{\|x\|^2}{s}&lt;t, s&gt;0, t&gt;0 \right\}$$   $-\ln(x)$ is a convex monotonic decreasing function, hence we can use composition by showing the inner part is concave. Also we know $-||x||^2$ is concave and $st$ is neither convex nor concave on $S$. Also in $S$ we have $st&gt;||x||^2$ but I was unable to show the convexity of the function as a whole
13060
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.   Now, I'm working on an exercise which essentialy asks to prove the following:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.   Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.  The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.  My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.  My idea to prove the claim was then:  First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. But since $\tilde{x}$ is the restriction of $x$ to $\mathbb{N}'$ and since $\mathbb{N}''\subset \mathbb{N}'$, the restriction of $\tilde{x}$ to $\mathbb{N}''$ is itself the restriction of $x$ to $\mathbb{N}''$.  In other words, every (sub)subsequence can be viewed as a subsequence of the original one. Considering this, by hypothesis every subsequence has a (sub)subsequence which converges to $a$. But by this argument, this means that we can replace every subsequence with a convergent one. And then we can use the result I stated.  Is my idea correct? Is this really the way to prove the claim? My only doubt is that I need every subsequence converging to $a$ to use the result. In this case I can't see how I can argue that this implies every subsequence converges to $a$.  How can I finish this proof?
70505
Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.  that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.  Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)  For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows: Let $\varepsilon&gt;0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)&lt;\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon&lt;(\mu_t\ast\nu_t)(K)&amp;=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{&lt;\:\varepsilon}\\&amp;&lt;\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)&gt;1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$  Can we drop the probability measure assumption?  EDIT I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon&gt;0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)&lt;\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)&lt;c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)&lt;k_3(\varepsilon)$$ The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$). In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)&lt;\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)  $^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon&gt;0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)&lt;\varepsilon.$$
90814
In  (section 2.3), it is stated that:     A point $x^*$ is a minimizer of a function $f$ (not necessarily convex) if and only if $f$ is subdifferentiable at $x^*$ and $0 \in\partial f(x^*).$   Could anybody provide me with references for a proof of the above statement?   Is there a reference where we can learn more about subgradients of non-convex functions?  In Section 3. Calculus of subgradients of the above notes,many properties of subgradients are presented for convex functions. I would like to know which ones among these properties still hold for non-convex functions.  Thank you in advance!
113001
$f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function. Assume that $\lim_{x \rightarrow \pm \infty} f(x)$ exist and are finite. Prove that $f$ is bounded.  So to show that $f$ is bounded, I must show that $\exists M \in \mathbb{R}$ such that $|f(x)| \leq M$ for all $x \in Dom(f)$. Assuming that $f(x)$'s limit exist, then I know that $\forall \epsilon  &gt; 0$, $\exists N$ such that for $n &gt; N$, $|f(n) - L| &lt; \epsilon$. But I am having trouble with figuring out how to connect the two ideas together.
10684
I read that in a complete Euclidean space - i.e. a normed real space with the norm induced by the scalar product - any sequence of nested bounded non-empty closed convex sets has a non-empty intersection, but I can't manage to prove it to myself. Has anybody any ideas or links to online proofs? Context The space is not assumed finite dimensional. At the point where I am in the text, Kolmogorov and Fomin's, they haven't defined a Hilbert space yet. I have not yet learned about the weak topology and reflexivity.
47059
I have tried to prove the following result, but it looks a bit messy in my opinion. How can it be improved?     Proposition. Let $a_n$ be a bounded sequence with values in $\mathbf{R}$.   Let $A=\{l\in\mathbf{R} \mid \text{ there exists a subsequence of $a_n$ &gt;converging to }l\}$. Then $A$ is closed.   Lemma: For each subsequence of $a_n$ which converges to $l$ it is true that $\forall {\varepsilon}&gt;0$ $a_n\in B(l,{\varepsilon})$ for arbitrarily large $n$ -- that is, for each $m\in\mathbf{N} \; \exists M\geq m \mid a_M\in B(l,{\varepsilon})$.  Let $a_{k_n}$\,be such a subsequence. By definition, for each $\varepsilon&gt;0 \; \exists N$ such that $\forall n\ge N \; a_{k_n}\in B(l,\varepsilon)$. Let $m\in\mathbf{N}$. Then $k_m\ge m$, since $a_{k_n}$ is a subsequence of $a_n$.   if $m\ge N$, take $M:=k_m$; if $m&lt;N$, take $M:=k_N$.   Proof. Let $x$ be a closure point of $A$. By definition, $\forall \varepsilon&gt;0 \; B(x,\varepsilon)\cap A\neq\varnothing$, which means that $\forall \varepsilon&gt;0$ there exists a subsequence $a_{k_n}$ which converges to ${l}_\varepsilon\in B(x,\varepsilon)$.  This means in particular that for all $m\in\mathbf{N}$ there exists a subsequence $a_{k^{(m)}_n}$ that converges to $l_m\in B(x,1/2^{m+1})$ .  Define the sequence $x_n$ inductively:   $x_0=0$; $x_n=\min\{m&gt;x_{n-1} \mid a_m \in B(l_m,1/2^{m+1}) \}$.    The sequence is well-defined, because there exists a subsequence converging to $l_m$, and by the lemma there are infinite indices for which the sequence lies in the neighborhood. So the set is not empty and by well-ordering there exists a minimum element.  It follows that for each $n\in\mathbf{N}$ $$x-1/2^{m}&lt;l_m-1/2^{m+1}&lt;a_{x_n}&lt;l_m+1/2^{m+1}&lt;x+1/2^{m},$$  so by squeeze theorem (strict inequalities imply non-strict inequalities) $a_{x_n}\to x$.
56970
I was wondering if this is true.     Let $f:(a,b) \rightarrow \mathbb{R}$ be a continuous real function and suppose $c \in (a,b)$ is the unique local maximum of $f$ and that $f$ has no other local extremum. Then $c$ is also an absolute maximum of $f$. The statement is analogous for minimums.    I have no idea how to prove this, nor have I thought of any counterexamples, though I am certain that the assumption of continuity is essential; otherwise consider something like $f:(0,\frac 32) \rightarrow \mathbb{R}$ given by  $$f(x) = \begin{cases} 1 - \frac x2 &amp;  x≤1 \\ \sqrt{\frac{1}{16} - (x-\frac 54)^2} &amp;  x&gt;1 \end{cases}$$  which admits a unique local maximum at $x=\frac 54$, no local minimum, but no absolute maximum.  Note that a similar question was asked  but the answers were not satisfactory and OP did not specify whether the interval was open or closed.
83662
Let $\{f_k\}_{k=1}^\infty\in L^{loc}_1(\mathbb R^n)$ be a sequence of real value functions such that  $\mbox{supp}(f_k)\subseteq\{|x|\leq k^{-1}\}$, $\quad \int f_k(x)dx=1$ $\quad for\quad all \quad k=1,2........$  show that the sequence $\{f_k^2\}_{k=1}^\infty$ does not converge in $\mathscr{D}^\prime(\mathbb R^n)$ $as \quad k\rightarrow \infty$
64952
Could you please give me a hint to prove that $\mathbb{R}^n$ with the 1-norm $\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert$ is not isometric to $\mathbb{R}^n$ with the infinity-norm $\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)$ for $n\gt2$.  I do not see how the critical case $n=2$ enters the picture.  Thank you!
113281
This seems to be a fairly easy question but I'm looking for new points of view on it and was wondering if anyone might be able to help.  (By the way- this question does come from home-work, but I've already solved and handed it, and I'm posting this out of interest, so no HW tag.)     Let $B_n=B(x_n,r_n)$ be a sequence of nested closed balls in a Banach space $X$. Prove that $\bigcap_{n=1}^\infty B_n\neq\varnothing$.   As I said before, it should be rather simple. When the radii decrease to 0, it's just a matter of selecting any sequence of points in $B_n$, and it must be Cauchy- and the limit is in the intersection.  My question is what to do when the radii do not decrease to 0? I got some tips about multiplying the balls by a sequence of decreasing scalars, or reducing the radii so that they decrease to 0, but found too many pathological cases for both methods.  Finally- I used a geometric arguemnt (which i've shown to work in any normed space) that if $B(x_1,r_1)\subset B(x_2,r_2)$ then $\| x_1-x_2\|\leq|r_1-r_2|$. This turned out to be some kind of technical catastrophe, but it worked...  Still, if anyone knows of a more elegant solution, I'd love to hear about it.
34793
In Section 2.5.2 of the book Convex Optimization by Boyd and Vandenberghe, the authors mentioned without proving that "a convex set in $\mathbb{R}^n$ with empty interior must lie in an affine set of dimension less than $n$." While I can intuitively understand this result, I was wondering how it can be proved formally?
54411
I am studying calculus of variation, and I need to prove that  $I[w] = \int_U \frac{1}{2} |Dw|^2 - fw \, dx$ with $f \in L^2(U)$  is weakly lower semicontinuous on $H_0^1(U)$.  In classes, I only learned that     Assume L is bounded below, and in addition $L(p,z,x)$ is convex in $p$,    for each $z \in \mathbb{R}$, $x \in U$. Then $I[.]$ is weakly lower semicontinuous on $W^{1,q}(U)$, where $I[w] = \int_U L(Dw, w, x)$.   And I think my $L$ isn't bounded below.   I don't know how I can solve the problem.