6 votes

Proof: How many continuous/bounded functions on $[0,1]$ verify $f(x)=f(x/2)\frac{1}{\sqrt{2}}$?

Let $M=\displaystyle\sup_{0\le x\le 1}|f(x)|.$ Then $$0\le M\le 2^{-1/2}\sup_{0\le x\le {1\over 2}}|f(x)|\le 2^{-1/2}M$$ Hence $M=0,$ i e. $f(x)= 0$ for any $x\in [0,1].$ Continuity is not essential. ...
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5 votes
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Is the map continuous?

Yes, you are approaching it correctly. Note that, in $[0,1]$, $[0,1)$ is an open set, since it is the ball centered at $0$ with radius $1$. By a similar argument, $(0,1]$ is an open subset of $[0,1]$, ...
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4 votes
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Continuity of polar decomposition

Indeed there is a very well developed spectral perturbation theory which tells you that two nearby operators have close spectra. Nevertheless, for the problem in hand, there is a much simpler ...
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  • 13.9k
4 votes
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How to find the primitive/ antiderivative of a discontinuous function

For sufficiently nice functions $f$ like the one in this example (e.g., functions that are continuous on some interval except perhaps at a finite number of jump discontinuities) and any $a$ in the ...
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4 votes
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Prove inequality $|f(x) - f(y)| \leq |x-y|$

You can just use algebra to prove. Note \begin{eqnarray} &&\left|\frac{1}{\sqrt{a^2 + 1}} - \frac{1}{\sqrt{b^2 + 1}}\right|\\ &=&\left|\frac{\sqrt{a^2 + 1}-\sqrt{b^2 + 1}}{\sqrt{a^2 + ...
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  • 35k
3 votes

Must a certain continuous map have 0 in its image, given that its restriction to the unit sphere is homotopic to the identity?

If, by contradiction, $f$ has no zero, then $f|_{\mathbb S^{n-1}}$ is homotopic to a constant in $\mathbb R^n\setminus\{0\}$ by $$ H(x,t)=f(tx), \quad x\in S^{n-1}, \ t\in[0,1]. $$ Therefore, by ...
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  • 13.9k
3 votes
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Proof: How many continuous/bounded functions on $[0,1]$ verify $f(x)=f(x/2)\frac{1}{\sqrt{2}}$?

Thank to @TonyK @Ryszard Szwarc and their comments. I think that i found an ever stronger demonstration that prooves that $\exists ! f_0(x)=0 \; s.t. \; f(x)=f(x/2)\frac{1}{\sqrt{2}}$ for all the ...
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3 votes
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Is $f :\mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}\times\mathbb{R},\ f((x_1,x_2),(y_1,y_2)):=(x_1,y_1)$ continuous?

The function is uniformly continuous, which by definition means that if the arguments are close to each other then the values are close to each other. In your case for two points $X=(x_1,x_2,y_1,y_2)$ ...
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2 votes

$f''(x)$ is always positive then $f(x+f'(x)) \geq f(x) $

As you recognized that $f(x)$ is a convex function, and it's differentiable. One property we can leverage is that $f(x)$ lives above the tangent line at any point $x$, which is $$ f(x+t)\geq f(x) + tf'...
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2 votes

How to find the primitive/ antiderivative of a discontinuous function

Antiderivative just refers to a function whose derivative will be this function. But since this function is discontinuous at $x=0$ the antiderivative you will compute will not be differentiable at $x=...
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2 votes

Proof: How many continuous/bounded functions on $[0,1]$ verify $f(x)=f(x/2)\frac{1}{\sqrt{2}}$?

Your proof seems ok especially idea of part III. However it is hard to read due to lack of some kind of order. It suffices to see that function $f$ has to be bounded (certainly it is) since is ...
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2 votes

What is range of values that the word 'nearby' supposed to represent in this informal definition of continuity.

I just want to know how near is nearby? It's as near as it needs to be. It should be seen as kind of a challenge. If you prescribe a distance $\varepsilon$ around $f(p)$, there is as distance $\delta$...
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2 votes
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How to prove that the convergence of $(f_n)$ is uniform on compact sets?

$\mathcal{G}$ locally equi-Lipschitz on $K$ $\implies$ $\mathcal{G}$ equi-Lipschitz on $K$: For each $x \in K$, there is a neighborhood $U_x$ and a constant $L_x \geq 0$ so that every function in the ...
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2 votes

Prove inequality $|f(x) - f(y)| \leq |x-y|$

Let $f(x) = \frac{1}{\sqrt{x^2+1}}$. We have to show $|f(x) - f(y)| \le |x - y|$. WLOG let $x \neq y$, then by MVT we know that $\exists c \in (x,y)$ s.t. $f'(c) = \frac{f(x) - f(y)}{x - y} $. So if ...
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  • 2,132
1 vote
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Is $\Lambda:X\longrightarrow\mathbb K$ continuous if $\{\Lambda(x_k)\}_{k\in\mathbb N}$ is bounded when $x_k \longrightarrow0$?

Suppose $\Lambda:X\longrightarrow\mathbb K$ is not continuous. Then $\forall n\in \Bbb{N}$ , $\exists (x_n) \in X$ such that $|\Lambda(x_n) |>n^2\|x_n\|$ Then $|\Lambda(\frac{x_n}{n\|x_n\|})|>n$ ...
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  • 4,246
1 vote
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Continuity of Taylor remainder for a multivariate $C^1$ function

Let $f:U\subset \mathbb{R}^n\to \mathbb{R}^m$ being a $\mathcal{C}^1$ function. Then $$f(u+h)=f(u)+f'(u)(h)+R(u,h),\qquad h\in\mathbb{R}^n, \quad u,u+h\in U,$$ in which $f'(u)$ is viewed as a linear ...
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1 vote

In this informal definition of continuity (at $x=p$), what does "regardless of the manner in which $x$ approaches $p$" mean?

I guess what they mean is that regardless of what sequence $\{x_n\}$ such that $\lim_{n\to\infty} x_n=p$ that you choose, you have that $\lim_{n\to \infty} f(x_n)= f(p)$. It just means that if $x$ is ...
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1 vote
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Example of discontinuous convex l.s.c. function on an open convex subset of an incomplete normed space

Let $X = c_c$ be the space of finite sequences, equipped with the $\ell^2$-norm. Then, $f \colon c_c \to \mathbb R$, $$ f(x) := \sum_{n = 1}^\infty n x_n^2 $$ is convex, lower semicontinuous (via ...
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  • 28k
1 vote

Showing that $f((x_1,x_2),(y_1,y_2)):=x_1-y_1$ is continuous.

Your idea is correct but your argumentation is a mess (there is no product going on here). First prove that the projection $\pi: \mathbb{R}^2 \to \mathbb{R}$ given by $\pi(x, y) = x$ is continuous. ...
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1 vote

Showing that $(1,\sqrt{2})\in U\subseteq(0,\infty)\times(0,\infty)\Longrightarrow \exists p,q\in\mathbb{N} :\Big(1,\frac{p}{q}\Big)\in U$

One issue: just because $U$ is an open subset of $(0, \infty) \times (0, \infty)$ does not mean you can write $$U = (a, b) \times (c, d).$$ However, since open sets of the form $(a, b) \times (c, d)$ ...
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