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We assume that we have an $\alpha$-Hölder continuous function $f$ on the interval $[0,1]$.

I am wondering if there exists an explicit construction of a sequence $f_{n} \in C_c^{\infty}(\mathbb R)$ such that

$$\lVert f-f_n\rVert_{C^{\beta}([0,1])} \le \frac{1}{n}$$

with $\beta<\alpha$ fixed and $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.

EDIT: My original question asked for $\beta=\alpha$ and I received two perfectly correct answers, but then Nik Weaver pointed out that the question is false for obvious reasons, independent of me requiring that $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$.

We assume that we have a $\alpha$-Hölder continuous function $f$ on an interval $[0,1]$ with $f(0)=0$.

I am wondering if there exists an explicit construction of a sequence $f_{n} \in C_c^{\infty}(\mathbb R)$ such that

$$\lVert f-f_n\rVert_{C^{\alpha}([0,1])} \le \frac{1}{n}$$

and $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.

We assume that we have an $\alpha$-Hölder continuous function $f$ on the interval $[0,1]$ with $f(0)=0$.

I am wondering if there exists an explicit construction of a sequence $f_{n} \in C_c^{\infty}(\mathbb R)$ such that

$$\lVert f-f_n\rVert_{C^{\beta}([0,1])} \le \frac{1}{n}$$

with $\beta<\alpha$ and $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.

EDIT: My original question asked for $\beta=\alpha$ and I received two perfectly correct answers, but then Nik Weaver pointed out that the question is ill-posed for obvious reasons.

We assume that we have an $\alpha$-Hölder continuous function $f$ on the interval $[0,1]$.

I am wondering if there exists an explicit construction of a sequence $f_{n} \in C_c^{\infty}(\mathbb R)$ such that

$$\lVert f-f_n\rVert_{C^{\beta}([0,1])} \le \frac{1}{n}$$

with $\beta<\alpha$ fixed and $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.

EDIT: My original question asked for $\beta=\alpha$ and I received two perfectly correct answers, but then Nik Weaver pointed out that the question is false for obvious reasons, independent of me requiring that $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$.

We assume that we have an $\alpha$-Hölder continuous function $f$ on the interval $[0,1]$ with $f(0)=0$.

I am wondering if there exists an explicit construction of a sequence $f_{n} \in C_c^{\infty}(\mathbb R)$ such that

$$\lVert f-f_n\rVert_{C^{\alpha}([0,1])} \le \frac{1}{n}$$

and $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.

We assume that we have an $\alpha$-Hölder continuous function $f$ on the interval $[0,1]$ with $f(0)=0$.

I am wondering if there exists an explicit construction of a sequence $f_{n} \in C_c^{\infty}(\mathbb R)$ such that

$$\lVert f-f_n\rVert_{C^{\beta}([0,1])} \le \frac{1}{n}$$

with $\beta<\alpha$ and $\lvert f_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.

EDIT: My original question asked for $\beta=\alpha$ and I received two perfectly correct answers, but then Nik Weaver pointed out that the question is ill-posed for obvious reasons.