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12

A vector space is just a set in which you can add and multiply by elements of the base field. You can add polynomials together and multiply them by real numbers (in a way satisfying the axioms,) so polynomials form a vector space. A vector is nothing more or less than an element of a vector space, so polynomials can be seen as vectors.


7

We want to mimic $f(x) = x^{-1/2}$ on $[0,1]$, but then cut $[0,1]$ into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller. So let $x_n \to 0$ be a decreasing sequence with $x_0 = 1$. Write $x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose ...


6

Do you remember how we can prove Cauchy-Schwarz through interpolation? Since $|xy|\leq\frac{x^2+y^2}{2}$ we have: $$ \| f\cdot g\|_1 \leq \frac{\|f\|_2^2}{2}+\frac{\|g\|_2^2}{2} \tag{1} $$ but the LHS is just the same if we replace $f$ with $\lambda f$ and $g$ with $\frac{1}{\lambda}g$, so: $$ \| f\cdot g\|_1 \leq ...


6

I think you are confused between the idea of a vector (ie. element of a set that forms a vector space) and the coordinate vectors of real numbers usually used to keep track of these vectors. More abstractly, vector spaces can be over any field. In the case of polynomials, after choosing a basis (this is always possible with finite dimension), we can write ...


6

A quick note: A 'space' is an arbitrary set of 'things' but with an additional structure or property. A metric space is a set where you can measure distances between two points, but you have to know the properties of the distance because they can get arbitrarily complicated (e.g. p-adic distance). So examples or explainations from others can help, but from ...


5

This is a case of equality in the Cauchy-Schwarz Inequality $$\left|\int_{0}^1 f\overline g\right|^2\leq \int_0^1 |f|^2\int _0^1|g|^2 $$ where $g=1$ is a constant function. And the equality holds if and only if $f$ and $g$ are dependent,i,e $f$ is a scalar multiple of $g$. Note that when we change the bounds to $a,b$, the equality in question is not ...


5

The formal definition of "vector space" follows a pattern that is seen very often in modern mathematics. You start with some particular object(s) - in the case of vector spaces these are the $n$-dimensional euclidean spaces - and figure out the essential laws governing these object(s). You then make these laws the axioms of some newly defined class of ...


5

Edited Answer : Let $O$ be an arbitrary point on the plane. $f(A)+f(p)+f(O)+f(S)+$$f(p)+f(O)+f(B)+f(q)+$$f(O)+f(q)+f(C)+f(r)+$$f(O)+f(s)+f(r)+f(D)=0$ This can be rearranged as $4f(O)+$$f(A)+f(B)+f(C)+f(D)+$$2(f(p)+f(q)+f(r)+f(s))$ $=4f(O)=0$.


4

I take it you are asking for compactness in $C^0$ Then no, of course not, in particular not in this general setup. You can choose $x_1 = -2, x_2 = +2 $, then all functions $(\sin(\frac{kt}{2\pi}))_{k\in\mathbb{N}} $ and $(\cos(\frac{kt}{2\pi}))_{k\in\mathbb{N}} $ are in that set and it is known that each continuous functions may be approximated by (linear ...


4

But they are vectors! Generally speaking, a space of functions mapping a topological space into a vector space satisfies the axioms of a vector space under the usual operations of addition and scalar multiplication. You don't need to have a basis to have a vector space, and in fact, it is often profitable to find or construct a basis that suits the problem ...


4

I would say it is almost impossible to get a good feeling of various kinds of spaces just by reading the definitions. Historically, those notions are not developed immediately by somebody in a dream or something like that. Instead, those concepts are formulated gradually. The concept of a manifold is a direct example. So, how these concepts are formulated? ...


4

I'm an engineering/physics student but I've also had to teach myself about certain types of spaces. I think the most important spaces to learn first to orient yourself are topological, metric, and vector spaces. Many spaces I've come across are special cases or combinations of these. Topological/metric spaces are more analytic (concerned with the ...


4

Since $L$ is linear, it suffices to show continuity at $0$. Assume $L$ is not continuous at $0$, and obtain from linearity that for any $a\in\mathbb{C}$ and any $\delta>0$, there is $x\in B(0,\delta)$ with $|L(x)|=|a|$, where $B(0,\delta)$ is the ball in $X$ with radius $\delta$, centered at $0$. Use linearity again to show that in fact, $x$ can be chosen ...


4

As $ T $ is bounded and self-adjoint with norm $ 1 $, its spectrum $ \sigma(T) $ is a compact subset of $ [-1,1] $, and its spectral radius $ r(T) $ equals $ 1 $. Hence, either $ -1 \in \sigma(T) $ or $ 1 \in \sigma(T) $. If $ -1 \in \sigma(T) $, then $ 0 \in \sigma(I + T) $, and so $ I + T $ is not invertible. If $ 1 \in \sigma(T) $, then $ 0 \in \sigma(I ...


4

Contrary to what I tried to do first, I think I now have a (complete) argument that your claim is false. Let me rewrite your last assumption (the displayed equation). Since $\int x^2|\widehat{g}(x)|^2 = \int |g'|^2$, this really says that certain derivatives are uniformly bounded in $L^2$ (originally, I may have to define these as distributional ...


4

1) Your function is clearly differentiable at $t$ if $t\ne 0$, since cosine and $t^2$ are differentiable. To show differentiability at $t=0$ use the limit-of-the-difference-quotient definition of the derivative, and use the fact that $|\gamma(t)|\le t^2$. This will quickly show that the derivative at $t=0$ is defined and is zero. 2) Consider that ...


4

It's equivalent to prove $T$ is continuous. One theorem to get $T$ is continuous is the closed graph theorem. So can you show the graph of ($x$,$Tx$) is closed? Suppose $x_n \rightarrow x$ and $Tx_n \rightarrow y$ and we want to show $Tx=y$. Indded, $\langle Tx, z \rangle = \langle x, T^*z \rangle =\lim_{n \to \infty} \langle x_n, T^*z \rangle = \lim_{n ...


4

The space of continuous functions on $[0,1]$ with the norm $\|x\|= \sup\{|x(t)| : t\in[0,1] \}$ is a Banach space. To justify that claim, you would need to know that all continuous functions on $[0,1]$ are bounded, and that Cauchy sequences in this metric space actually converge, and those take some work. One way to prove that this norm does not come from ...


4

There are lots of ways to generalize the derivative, not all of which are compatible. This is a common feature of generalizations: for example, there are also lots of ways to generalize numbers, or infinity, or exponentiation, not all of which are compatible. Here are some generalizations of derivatives you can write down: The Fréchet derivative of a map ...


4

Take $g(x) = x^2$. Consider $f_n \to 0$ in $L^2$. If your property was true, it would means that $f_n \to 0$ in $L^4$ Now just take $$f_n(x) = x^{-\frac{1}{4}} \times\mathbb{1}_{[0,\frac{1}{n}]}(x)$$ $f_n \to 0$ in $L^2$, but $f_n$ is not in $L^4$ Hence you have both $$\int_0^1 [f_n(x)- 0|^2 dx = \int_0^{\frac{1}{n}} \frac{1}{\sqrt{x}} dx = ...


4

This is part of the point of Hölder's inequality. It shows that if $f\in L^p$, $g\in L^q$ and $\frac1p+\frac1q=1$ then $fg\in L^1$. To see "why" this holds, I suggest you examine a proof of Hölder's inequality. I can provide one if you do not have one. EDIT: The proof relies on Young's inequality: if $x,y\ge0$ then $xy\le\frac{x^p}p+\frac{y^q}q$. This ...


4

I can answer this question when the cardinality $\lambda$ of Hamel basis of $X$ is not less than $\mathfrak{c}$. For any infinite dimensional Banach space its cardinality and cardinality of its Hamel basis are equal. See theorem 3.5 from The cardinality of Hamel bases of Banach spaces by Lorenz Halbeisen , Norbert Hungerbühler. Consider Banach space ...


3

No, $\sin n$ does not work. You can use the Lorenz criterion (see Martin's answer). In fact, if $a_n = \sin \log n$ were almost convergent (to value $L$), then in particular $$ \lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n \sin \log k = L $$ But $$ \frac{1}{n}\sum_{k=1}^n \sin \log k - \frac{1}{n}\int_1^{n+1}\sin\log x\;dx $$ goes to zero, so we may show the ...


3

Note that specifying a pair of linear functionals on $X$ is the same as specifying a single linear functional $X\oplus X$, by writing $M(x\oplus y)= L_1(x)\oplus L_2(y)$. Also, if $p,q$ are sublinear functionals on $X$ then $r(x\oplus y)=p(x)+q(y)$ defines a sublinear functional on $X\oplus X$. Consider now applying Hahn-Banach to extend the functional $M$ ...


3

Use the fundamental theorem and apply Holder's inequality: for $x > 0$ you have $$|f(x)| = |f(x) - f(0)| \le \int_0^x |f'(t)| \, dt \le \left(\int_0^x \, dt\right)^{2/3} \left( \int_0^x |f'(t)|^3 \, dt\right)^{1/3}$$ so that $$ |x^{-2/3} f(x)| \le \left( \int_0^x |f'(t)|^3 \, dt \right)^{1/3},\quad x > 0.$$ But $f' \in L^3$ implies $$ \lim_{x \to 0^+} ...


3

Since $R^2$ is finite dimensional, any norm on $R^2$ is complete. You only need to check whether $||\cdot||_1$ is a norm. As Tyler pointed, $||\cdot||_1$ does not satisfy the parallelogram law , so this can't naturally induce an inner product.


3

We have $$\hat{f}\left(\xi\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi i\xi x}dx$$ then $$\hat{f}\left(0\right)=\int_{-\infty}^{\infty}f\left(x\right)dx=\sum_{k\in\mathbb{Z}}\int_{k}^{k+1}f\left(x\right)dx=\sum_{k\in\mathbb{Z}}\int_{0}^{1}f\left(x-k\right)dx=\int_{0}^{1}\sum_{k\in\mathbb{Z}}f\left(x-k\right)dx=1.$$


3

If $n > 1$, then $$\int_3^{\infty} \frac1{x^n \ln x} \, dx < \int_3^{\infty} \frac1{x^n } \, dx = \frac1{n-1}3^{1-n}< \infty.$$ If $n = 1$, then $$\int_3^{\infty} \frac1{x \ln x} \, dx =\left.\ln(\ln x)\right|_3^{\infty} =\infty.$$ Can you figure out the case where $n < 1$?



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