# Tag Info

### 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. ...
Accepted

### 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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### 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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### 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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### 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 + ...

### 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 ...
Accepted

### 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 ...
### 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 $X=(x_1,x_2,y_1,y_2)$ ...